Answer:
- 431.15 kJ/mol.
Explanation:
Firstly, we can calculate the amount of heat (Q) released by the solution using the relation:Q = m.c.ΔT,
where, Q is the amount of heat released from the solution (Q = ??? J).
m is the mass of solution (m = 1.5 g + 100 g = 101.5 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 33.1°C - 22°C = 11.1°C).
∴ Q = m.c.ΔT = (101.5 g)(4.18 J/g.°C)(11.1°C) = 4709.4 J.
To find ΔH:∵ ΔH = Q/n
no. of moles of Ba (n) = mass/atomic mass = (1.50 g)/(137.3270 g/mol) = 0.011 mol.
∴ ΔH = - Q/n = (4709.4 J)/(0.011 mol) = - 431.15 kJ/mol.
The negative sign is not from calculation, but it is an indication that the reaction is exothermic.
The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 content in a normal living piece of wood. If the half-life of carbon-14 is 5,730 years, how old is the harpoon handle?
Answer:
3,964 years.
Explanation:
It is known that the decay of a radioactive isotope isotope obeys first order kinetics.Half-life time is the time needed for the reactants to be in its half concentration.If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).Also, it is clear that in first order decay the half-life time is independent of the initial concentration.The half-life of the element is 5,730 years.For, first order reactions:k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
Also, we have the integral law of first order reaction:kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
The wooden harpoon handle found in the Inuit archaeological site, with 61.9% of the original carbon-14 content remaining and knowing that the half-life of carbon-14 is 5,730 years, is approximately 4000 years old.
Explanation:To determine the age of the wooden harpoon handle found in an Inuit archaeological site, we use the concept of carbon-14 dating. Since the half-life of carbon-14 is 5,730 years, this means that after 5,730 years, 50% of the original carbon-14 would have decayed. In our case, the handle has 61.9% of the original carbon-14 content remaining.
We can formulate the decay process mathematically using the equation N(t) = N0 * (1/2)^(t/T), where N(t) is the remaining amount of carbon-14 at time t, N0 is the initial amount of carbon-14, T is the half-life, and t is the time that has passed.
Applying this to our case, we're looking to find 't', given that N(t)/N0 = 0.619 and T = 5730 years. The equation becomes 0.619 = (1/2)^(t/5730).
To solve for 't', we take the natural logarithm of both sides: ln(0.619) = ln((1/2)^(t/5730)) => ln(0.619) = (t/5730)*ln(1/2). Solving for 't', we find that 't' = 5730 * ln(0.619) / ln(1/2). This yields 't' = 5,730 * (-0.4797) / (-0.6931) = approximately 4000 years.
Therefore, the wooden harpoon handle is approximately 4000 years old based on its carbon-14 content.
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At a given temperature, 4.06 atm of H2 and 3.5 atm of Cl2 are mixed and allowed to come to equilibrium. The equilibrium pressure of HCl is found to be 1.418 atm. Calculate Kp for the reaction at this temperature. H2(g) + Cl2(g) <=> 2 HCl(g)
Kp calculated using equilibrium partial pressures of reactants and products is 0.206198. An ICE table helps determine the changes in pressures of H₂, Cl₂, and HCl during the reaction. After finding the equilibrium pressures, the Kp value is determined using the equilibrium constant expression.
To calculate the equilibrium constant Kp for the reaction H₂(g) + Cl₂(g) ⇌ 2 HCl(g), we can use the equilibrium partial pressures of the reactants and products. Based on the reaction stoichiometry, the change in pressure for H₂ and Cl₂ should be equal and opposite to the change for HCl, since two moles of HCl are produced for each mole of H₂ and Cl₂ that react.
Using the equilibrium pressure of HCl (1.418 atm) and the initial pressures of H₂ (4.06 atm) and Cl₂ (3.5 atm), we can set up an ICE table to find the changes in pressure (Δ) during the reaction and then the equilibrium pressures of H₂ and Cl₂.
Let x represent the change in pressure for H₂ and Cl₂. Since the ratio of HCl to H₂ and Cl₂ in the balanced equation is 2:1, the change in pressure for HCl will be 2x. If 1.418 atm is the equilibrium pressure of HCl, the change is the same amount (since HCl starts at 0 atm), and thus x is half of this value, which is 0.709 atm.
We can now calculate the equilibrium pressures of H₂ and Cl₂ by subtracting x from their initial pressures:
P(H₂) = 4.06 atm - x = 4.06 atm - 0.709 atm = 3.351 atmP(Cl₂) = 3.5 atm - x = 3.5 atm - 0.709 atm = 2.791 atmAfter determining the equilibrium pressures for all species, we apply the equilibrium expression for the reaction to find Kp:
Kp = [P(HCl)] ²/ [P(H₂) * P(Cl₂)]
Kp = [1.418]² / [3.351 * 2.791]
Kp = [2.010724] / [3.351 * 2.791]
Kp = [2.010724] / [9.75141]
Kp = 0.206198
By calculating this expression we get the value of Kp for the reaction at the given temperature is 0.206198.
Typical residential shower controls mix streams of hot water (140 F, or 60 C) with cold water (60 F, or 15 C) to form a stream of 40 C (104 F) water. The entire system loses energy to the surroundings at a rate of 5 kJ/kg of exiting water. What is the ratio of cold water-to-hot water mass flow rates necessary to provide the 40 C water?
Answer:
4:5
Explanation:
Let x represent the fraction of the mix that is hot water. Then the temperature of the mix is ...
60x +15(1-x) = 40·1
45x = 25 . . . . . . . . . subtract 15
x = 25/45 = 5/9 . . . divide by the coefficient of x
This is the fraction that is hot water, so the fraction that is cold water is ...
1-5/9 = 4/9
The ratio of cold to hot is ...
cold : hot = (4/9) : (5/9) = 4 : 5
_____
Additional comments
The problem assumes that the energy contained in a given mass of water is proportional to its temperature. That is almost true, sufficiently so that we can reasonably use that approximation.
If heat loss is figured into the problem, then additional information is needed regarding the energy content of water at temperatures in the range of interest. That is not provided by this problem statement, so we have ignored the heat loss.
Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one species to another. Oxidation states, or oxidation numbers, allow chemists to keep track of these electron transfers. In general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number. The species that is oxidized is called the reducing agent or reductant. The species that is reduced is called the oxidizing agent or oxidant. To sum up: Oxidation = increase in oxidation state = loss of electrons = reducing agent Reduction = decrease in oxidation state = gain of electrons = oxidizing agent Part A Which element is oxidized in this reaction? Fe2O3+3CO→2Fe+3CO2 Enter the elemental symbol. View Available Hint(s) is oxidized Part B Which element is reduced in this reaction? 2HCl+2KMnO4+3H2C2O4→6CO2+2MnO2+2KCl+4H2O Enter the elemental symbol. View Available Hint(s) is reduced
Fe2O3+3CO→2Fe+3CO2, 3CO is oxidized in this reaction, as oxygen is added to it. 2HCl+2KMnO4+3H2C2O4→6CO2+2MnO2+2KCl+4H2O, 3H2C2O4 is reduced as hydrogen is added and 2KMnO4 is reduced as oxygen is removed.
What is oxidation?Oxidation is defined as the process in which loss of electrons take place during the reaction by a molecule atom or ion.
Oxidizing agent is defined as a substance that causes oxidation by accepting electron therefore it gat reduced.
Reduction is defined as the transfer of electrons between spices in a chemical reaction. It shows loss of oxygen.
Reducing agent is defined as one of the reactant of redox reaction which reduces the another reactant by giving out electrons to the reactant.
Thus, Fe2O3+3CO→2Fe+3CO2, 3CO is oxidized in this reaction, as oxygen is added to it. 2HCl+2KMnO4+3H2C2O4→6CO2+2MnO2+2KCl+4H2O, 3H2C2O4 is reduced as hydrogen is added and 2KMnO4 is reduced as oxygen is removed.
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A cylinder, with a piston pressing down with a constant pressure, is filled with 2.00 moles of a gas (n1), and its volume is 50.0 L (V1). If 0.400 mole of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?
Answer:
40.0 L.
Explanation:
We can use the general law of ideal gas: PV = nRT.where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If P and T are constant, and have different values of n and V:(V₁n₂) = (V₂n₁).
V₁ = 50.0 L, n₁ = 2.0 moles,
V₂ = ??? L, n₂ = 2.0 mol - 0.4 mol = 1.6 mol.
∴ V₂ = (V₁n₂)/(n₁) = (50.0 L)(1.6 mol)/(2.0 mol) = 40.0 L.
) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79% N2) is fed to provide 35% excess oxygen, and the fractional conversion of HCl is 85%. Calculate the mole fractions of the product stream components.
Answer:
Here's what I get.
Explanation:
1. Write the chemical equation
[tex]\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O[/tex]
Assume that we start with 4 L of HCl
2. Calculate the theoretical volume of oxygen
[tex]\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}[/tex]
3. Add 35% excess
[tex]\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}[/tex]
4. Calculate the theoretical volume of nitrogen
[tex]\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}[/tex]
4. Calculate volumes of reactant used up
Only 85 % of the HCl is converted.
We can summarize the volumes in an ICE table
4HCl + O₂ + N₂ → 2Cl₂ + 2H₂O
I/L: 4 1.35 5.08 0 0
C/L: -0.85(4) -0.85(1) 0 +0.85(2) +0.85(2)
E/L: 0.60 0.50 5.08 1.70 1.70
5. Calculate the mole fractions of each gas in the product stream
Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L
[tex]\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\[/tex]
The mole fractions of the product stream components are as follows:
HCl | 0.12
Cl2 | 0.50
H2O | 0.23
N2 | 0.79
Calculation of mole fractions of the product stream components using atomic species balances:
Step 1: Write the balanced chemical equation for the Deacon process:
2 HCl + O2 → Cl2 + H2O
Step 2: Define the atomic species and their mole fractions in the feed and product streams:
Feed stream:
* HCl: mole fraction = x_HCl
* O2: mole fraction = x_O2 = 0.21
* N2: mole fraction = x_N2 = 0.79
Product stream:
* HCl: mole fraction = y_HCl
* Cl2: mole fraction = y_Cl2
* H2O: mole fraction = y_H2O
* N2: mole fraction = y_N2
Step 3: Write the atomic species balances:
* Hydrogen:
2 x_HCl_feed = y_HCl_product + y_H2O_product
* Chlorine:
2 x_HCl_feed = y_Cl2_product
* Oxygen:
x_O2_feed + 0.35 x_O2_feed = y_O2_product + y_H2O_product
* Nitrogen:
x_N2_feed = y_N2_product
Step 4: Solve the atomic species balances:
We can solve the atomic species balances to obtain the following expressions for the mole fractions of the product stream components:
y_HCl_product = (2 x_HCl_feed - y_H2O_product) / 2
y_Cl2_product = 2 x_HCl_feed
y_H2O_product = 0.65 x_O2_feed - (2 x_HCl_feed - y_H2O_product)
y_N2_product = x_N2_feed
Step 5: Substitute the known values and solve for the mole fractions of the product stream components:
We know that the fractional conversion of HCl is 85%. This means that 85% of the HCl in the feed stream is converted to Cl2 and H2O. Therefore, the mole fraction of HCl in the product stream is:
y_HCl_product = 0.15 x_HCl_feed
We also know that the feed stream contains 21 mole% O2 and 79 mole% N2. Therefore, the mole fractions of O2 and N2 in the feed stream are:
x_O2_feed = 0.21
x_N2_feed = 0.79
Now we can substitute all of the known values into the equations above to solve for the mole fractions of the product stream components:
y_HCl_product = 0.15 x_HCl_feed
y_Cl2_product = 2 x_HCl_feed
y_H2O_product = 0.65 x_O2_feed - (2 x_HCl_feed - y_H2O_product)
y_N2_product = x_N2_feed
Solving these equations, we obtain the following mole fractions of the product stream components:
y_HCl_product = 0.12
y_Cl2_product = 0.50
y_H2O_product = 0.23
y_N2_product = 0.79
Therefore, the mole fractions of the product stream components are as follows:
**Component** | **Mole fraction**
------- | --------
HCl | 0.12
Cl2 | 0.50
H2O | 0.23
N2 | 0.79
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Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vessel is charged with a mixture of PCl3(g) and Cl2(g), which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are PPCl3 = 0.125atm , PCl2 = 0.155atm , and PPCl5 = 1.90atm Kp= 98.1 What is Kc?
Answer:
The equilibrium constant in terms of concentration that is, [tex]K_c=3.6243\times 10^{3}[/tex] .
Explanation:
[tex]PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)[/tex]
The relation of [tex]K_c\& K_p[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
[tex]K_p[/tex]= Equilibrium constant in terms of partial pressure.=98.1
[tex]K_c[/tex]= Equilibrium constant in terms of concentration =?
T = temperature at which the equilibrium reaction is taking place.
R = universal gas constant
[tex]\Delta n_g[/tex] = Difference between gaseous moles on product side and reactant side=[tex]n_{g,p}-n_{g.r}=1-2=-1[/tex]
[tex]98.1=K_c(RT)^{-1}[/tex]
[tex]98.1 =\frac{K_c}{RT}[/tex]
[tex]K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3} [/tex]
The equilibrium constant in terms of concentration that is, [tex]K_c=3.6243\times 10^{3}[/tex] .
The Kc for the reaction is obtained as 3620.
We have to apply the formula;
Kp =Kc(RT)^Δng
Where;
Kp = 98.1
Kc = ?
R= 0.082 LatmK-1mol-1
T = 450 K
Δng = 1 - 2 = -1
Now we have to substitute into the equation;
98.1 = Kc(0.082 × 450)^-1
98.1 = Kc/(0.082 × 450)
Kc = 98.1 (0.082 × 450)
Kc = 3620
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Draw the products obtained from the reaction of 1 equivalent of HBr with 1 equivalent of 2,5-dimethyl-1,3,5-hexatriene. Draw the molecules on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms, and Templates toolbars
Answer:
Here's what I get.
Explanation:
According to Markovnikov's rule, the H will add to a terminal carbon, generating three resonance stabilized carbocations.
The Br⁻ ion will add to any of the three carbocations.
There are three possible products:
5-bromo-2,5-dimethylhexa-1,3-triene (1) 3-bromo-2,5-dimethylhexa-1,4-triene (2) 1-bromo-2,5-dimethylhexa-2,4-triene (3)Copper reacts with silver nitrate through single replacement. If 2.75 g of silver are produced from the reacti?If 2.75 g of silver are produced from the reaction, how many moles of copper(II) nitrate are also produced? How many moles of each reactant are required in this reaction?
Answer:
[tex]\boxed{\text{0.0128 mol Cu(NO$_{3}$)$_{2}$; 0.0128 mol Cu; 0.0225 mol AgNO$_{3}$}}[/tex]
Explanation:
a) Balanced equation
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 63.55 169.87 187.56 107.87
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂ + 2Ag
m/g: 2.75
(b) Moles of Cu(NO₃)₂
(i) Calculate the moles of Ag
[tex]n = \text{2.75 g Ag} \times \dfrac{\text{1 mol Ag}}{\text{107.8 g Ag}} = \text{0.025 51 mol Ag}[/tex]
(ii) Calculate the moles of Cu(NO₃)₂
The molar ratio is 1 mol Cu(NO₃)₂:2 mol Ag
[tex]n = \text{0.025 51 mol Ag}\times \dfrac{\text{1 mol Cu(NO$_{3}$)$_{2}$}}{\text{2 mol Ag}} = \boxed{\textbf{0.0128 mol Cu(NO$_{3}$)$_{2}$}}[/tex]
(c) Moles of Cu
The molar ratio is 1 mol Cu:2 mol Ag
[tex]n = \text{0.025 51mol Ag} \times \dfrac{\text{1 mol Cu}}{\text{2 mol Ag}}= \boxed{\textbf{0.0128 mol Cu}}[/tex]
(d) Moles of AgNO₃
The molar ratio is 2 mol Ag:2 mol AgNO₃
[tex]n = \text{0.025 51 mol Ag} \times \dfrac{\text{2 mol AgNO$_{3}$}}{\text{2 mol Ag}}= \boxed{\textbf{0.0255 mol AgNO$_{3}$}}[/tex]
Which combination will produce a precipitate? Which combination will produce a precipitate? KOH (aq) and HNO3 (aq) AgC2H3O2 (aq) and HC2H3O2 (aq) Pb(NO3)2 (aq) and HCl (aq) Cu(NO3)2 (aq) and KC2H3O2 (aq) NaOH (aq) and Sr(NO3)2 (aq)
Explanation:
A precipitate is defined as an insoluble substance that emerges upon mixing of two aqueous solutions.
For example, [tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]
As precipitate is a solid and it is represented by (s). And, an aqueous solution is represented by (aq).
So, the products of the given reactants will be as follows.
[tex]KOH(aq) + HNO_{3} \rightarrow KNO_{3}(aq) + H_{2}O(aq)[/tex]
[tex]AgC_{2}H_{3}O_{2}(aq) + HC_{2}H_{3}O_{2}(aq) \rightarrow \text{No reaction}[/tex]
[tex]Pb(NO_{3})_{2}(aq) + HCl(aq) \rightarrow PbCl_{2}(s) + 2HNO_{3}(aq)[/tex]
[tex]Cu(NO)_{3}_{2}(aq) + CH_{3}COOK(aq) \rightarrow Cu(CH_{3}COO)_{2})(aq) + 2KNO_{3}(aq)[/tex]
[tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]
Hence, we can conclude that out of the given options, these two equations will produce a precipitate.
[tex]Pb(NO_{3})_{2}(aq) + HCl(aq) \rightarrow PbCl_{2}(s) + 2HNO_{3}(aq)[/tex]
[tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]
The combinations of Pb(NO3)2 (aq) and HCl (aq), and NaOH (aq) and Sr(NO3)2 (aq) will produce precipitates, as the resultant compounds (PbCl2 and Sr(OH)2 respectively) are insoluble in solutions.
Explanation:In the context of chemistry, a precipitate is a solid that forms in a solution during a chemical reaction. The combination that will produce a precipitate can be predicted using solubility rules, which detail the solubility of different compounds.
Between KOH (aq) and HNO3 (aq), no precipitate forms because the compounds formed are soluble in water. For the combination of AgC2H3O2 (aq) and HC2H3O2 (aq), no precipitate is expected as all resulting combinations are soluble. When Pb(NO3)2 (aq) and HCl (aq) are mixed, a precipitate of PbCl2 forms indicating a precipitation reaction. Pb²+ (aq) + Cl- (aq) → PbCl2 (s) Combining Cu(NO3)2 (aq) and KC2H3O2 (aq), no precipitate is expected as all potential combinations are soluble. Finally, the combination of NaOH (aq) and Sr(NO3)2 (aq) will lead to the formation of Sr(OH)2, an insoluble compound resulting in a precipitation reaction.Learn more about Chemical Reaction here:
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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: CH4(g)+H2O(g)→CO(g)+3H2(g) In a particular reaction, 26.0 L of methane gas (measured at a pressure of 734 torr and a temperature of 25 ∘C) is mixed with 23.0 L of water vapor (measured at a pressure of 700 torr and a temperature of 125 ∘C). The reaction produces 26.0 L of hydrogen gas measured at STP. Part A What is the percent yield of the reaction? %
Answer:
60.42% is the percent yield of the reaction.
Explanation:
Moles of methane gas at 734 Torr and a temperature of 25 °C.
Volume of methane gas = V = 26.0 L
Pressure of the methane gas = P = 734 Torr = 0.9542 atm
Temperature of the methane gas = T = 25 °C = 298.15 K
Moles of methane gas = n
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol[/tex]
Moles of water vapors at 700 Torr and a temperature of 125 °C.
Volume of water vapor = V' = 23.0 L
Pressure of water vapor = P' = 700 Torr = 0.9100 atm
Temperature of water vapor = T' = 125 °C = 398.15 K
Moles of water vapor gas = n'
[tex]P'V'=n'RT'[/tex]
[tex]n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol[/tex]
[tex]CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)[/tex]
According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.
According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.
Then 0.6402 moles of water vapor will give:
[tex]\frac{3}{1}\times 0.6402 mol=1.9208 mol[/tex] of hydrogen gas
Moles of hydrogen gas obtained theoretically = 1.9208 mol
The reaction produces 26.0 L of hydrogen gas measured at STP.
At STP, 1 mole of gas occupies 22.4 L of volume.
Then 26 L of volume of gas will be occupied by:
[tex]\frac{1}{22.4 L}\times 26 L= 1.1607 mol[/tex]
Moles of hydrogen gas obtained experimentally = 1.1607 mol
Percentage yield of hydrogen gas of the reaction:
[tex]\frac{Experimental}{Theoretical}\times 100[/tex]
[tex]\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%[/tex]
60.42% is the percent yield of the reaction.
The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00 L of H2O that is in contact with pure O2 at 1.00 atm? g O2 (b) How many grams of O2 will dissolve in 4.00 L of H2O that is in contact with air where the partial pressure of O2 is 0.209 atm?
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = 3.45 x 10⁻² grams O₂(g) in 4L water.
0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.
0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.
Explanation:
Henry's law states that the solubility of a gas in a given volume of a liquid is directly proportional to the partial pressure of that gas above the liquid.[tex]S=k_H\times p_o[/tex]
Where:
S = Solubility of gas in liquid
[tex]k_H[/tex]= Henry's law constant
[tex]p_o[/tex]= Partial pressure of a gas
Given:
The Henry's law constant for oxygen gas in water at 20°C is [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
To find:
a) Mass of oxygen gas in 4.00 Liter of water with pure oxygen at 1.00 atm.
b) Mass of oxygen gas in 4.00 Liter of water with oxygen gas at a partial pressure of 0.209 atm.
Solution:
a)
The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
The pressure of pure oxygen gas above water = [tex]p_o=1.00 atm[/tex]
The solubility of the oxygen gas in water:
[tex]S=1.28\times 10^{-3} mol/(Latm)\times 1.00 atm\\S=1.28\times 10^{-3} mol/L[/tex]
There are [tex]1.28\times 10^{-3}[/tex] moles of oxygen gas in 1 liter of water
The volume of the water = 4.00 L
Moles of oxygen gas in 4.00 L of water:
[tex]=1.28\times 10^{-3}\times 4.00 mol=5.12\times 10^{-3}mol[/tex]
Mass of [tex]5.12\times 10^{-3}[/tex] moles of oxygen gas:
[tex]=5.12\times 10^{-3}mol\times 31.998 g/mol=0.164 g[/tex]
0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.
b)
The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]
The partial pressure of oxygen gas above water = [tex]p_o=0.209atm[/tex]
The solubility of the oxygen gas in water:
[tex]S=1.28\times 10^{-3} mol/(Latm)\times 0.209atm\\S=2.68\times 10^{-4} mol/L[/tex]
There are [tex]2.68\times 10^{-4}[/tex] moles of oxygen gas in 1 liter of water
The volume of the water = 4.00 L
Moles of oxygen gas in 4.00 L of water:
[tex]=2.68\times 10^{-4}\times 4.00 mol=1.072\times 10^{-3}mol[/tex]
Mass of [tex]1.072\times 10^{-3}[/tex] moles of oxygen gas:
[tex]=1.072\times 10^{-3}mol\times 31.998 g/mol=0.0343g[/tex]
0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.
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Which of the following reactions could be an elementary reaction? 2 NO2(g) + F2(g) → 2NO2F(g) Rate = k[NO2][F2] H2(g) + Br2(g) → 2 HBr(g) Rate = k[H2][Br2]1/2 NO(g) + O2(g) → NO2(g) + O(g) Rate = k[NO][O2] NO2(g) + CO(g) → NO(g) + CO2(g) Rate = k[NO2]2
Answer: The correct answer is [tex]NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2][/tex]
Explanation:
Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
For the given reactions:
Equation 1: [tex]2NO_2(g)+F_2(g)\rightarrow 2NO_2F(g);Rate=k[NO_2][F_2][/tex]Molecularity of the reaction = 2 + 1 = 3
Order of the reaction = 1 + 1 = 2
This is not considered as an elementary reaction.
Equation 2: [tex]H_2(g)+Br_2(g)\rightarrow 2HBr(g);Rate=k[H_2][Br_2]^{1/2}[/tex]Molecularity of the reaction = 1 + 1 = 2
Order of the reaction = [tex]1+\frac{1}{2}=\frac{3}{2}[/tex]
This is not considered as an elementary reaction.
Equation 3: [tex]NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2][/tex]Molecularity of the reaction = 1 + 1 = 2
Order of the reaction = 1 + 1 = 2
This is considered as an elementary reaction.
Equation 4: [tex]NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g);Rate=k[NO_2]^2[/tex]Molecularity of the reaction = 1 + 1 = 2
Order of the reaction = 2 + 0 = 2
In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction. Hence, this is not considered as an elementary reaction.
Hence, the correct answer is [tex]NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2][/tex]
An elementary reaction is a reaction that occurs in a single step and involves the collision of reactant molecules. In this case, both of the given reactions could be considered elementary.
Explanation:An elementary reaction is a reaction that occurs in a single step and involves the collision of reactant molecules. To determine if a reaction is elementary, we need to examine the overall reaction and see if it can be written as a sum of individual elementary reactions.
In this case, the reaction 2 NO2(g) + F2(g) → 2NO2F(g) can be written as a sum of individual elementary reactions:
NO2(g) + F2(g) → NO2F(g)
NO2(g) + F2(g) → NO(g) + FNO(g)
Since the overall reaction can be broken down into individual elementary reactions, both of these reactions could be considered elementary.
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Consider the reaction: 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) ΔH = -198.2 kJ/mol. What would be the effect on the reaction if we were to:a. increase the temperatureb. increase the pressurec. increase [SO2]d. add a catalyst
Answer:
a. increase the temperature: the reaction will be shifted to the left side.
b. increase the pressure: the reaction will be shifted to the right side.
c. increase [SO₂]: so, the reaction will be shifted to the right side.
d. add a catalyst: it has no effect.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
a. increase the temperature:
∵ ΔH has a negative value, the reaction is exothermic.
The heat can be represented as a part of the products.Increase the T will increase the concentration of the products "heat".So, the reaction will be shifted to the left side to suppress the increase in the concentration of products by increasing T.So, the reaction will be shifted to the left side.
b. increase the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.The reactants side (left) has 3.0 moles of gases and the products side (right) has 2.0 moles of gases.Thus, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).So, the reaction will be shifted to the right side.
c. increase [SO₂]:
Increasing [SO₂] will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of SO₂.So, the reaction will be shifted to the right side.
d. add a catalyst:
Catalyst increases the rate of the reaction without affecting the equilibrium position.Catalyst increases the rate via lowering the activation energy of the reaction.This can occur via passing the reaction in alternative pathway (changing the mechanism).The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.So, it has no effect.
The effect on the SO2 + O2 ⇌ SO3 reaction based on Le Chatelier's Principle: increasing temperature will favor SO2 and O2 formation; increasing pressure will favor SO3 formation; increasing SO2 concentration will favor SO3 formation; adding a catalyst won't shift the equilibrium but will help attain it quicker.
Explanation:Considering the reaction: 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) ΔH = -198.2 kJ/mol in relation to Le Chatelier's Principle, the following outcomes can be expected:
a. Increase the temperature: This reaction is exothermic, where heat is also considered a product. Increasing the temperature will shift the equilibrium to the left, favoring the reverse reaction and producing more reactants (SO2 and O2).b. Increase the pressure: By increasing the pressure, the equilibrium will shift to the side with fewer gas molecules. In this case, the equilibrium will shift to the right, resulting in more SO3.c. Increase [SO2]: Increasing the concentration of SO2 will shift the equilibrium to the right, favoring the forward reaction and producing more SO3.d. Add a catalyst: The addition of a catalyst will speed up both the forward and reverse reactions equally. It will not shift the equilibrium but will help the system reach equilibrium faster.Learn more about Le Chatelier's Principle here:https://brainly.com/question/33440831
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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and stirred until it dissolves. What is the final temperature of the solution? ΔHsoln =-44.51 kJ/mol.
Answer:
37.1°C.
Explanation:
Firstly, we need to calculate the amount of heat (Q) released through this reaction:
∵ ΔHsoln = Q/n
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
The negative sign of ΔHsoln indicates that the reaction is exothermic.
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
We can use the relation:Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
∴ final temperature = 25°C + 12.1 = 37.1°C.
The values for the enthalpies of formation of C6H6(l), CO2(g) and H2O(l) are 49 kJ mol-1, -393 kJ mol-1, and -285 kJ mol-1 respectively. Use this information to determine the enthalpy change when 1 mole of C6H6(l) undergoes combustion to produce carbon dioxide gas and liquid water.
Answer:
ΔH(Combustion C₆H₆ = -3,262 Kj/mole
Explanation:
It takes one oxygen to combine with two hydrogen atoms to form water. What is the mass ratio of oxygen to hydrogen? 1 to 2 4 to 1 8 to 1 2 to 1
Answer:
8 to 1.
Explanation:
Oxygen combines with hydrogen atoms to form water according to the balanced equation:O₂ + 2H₂ → 2H₂O.
It is clear that one mole of oxygen combines with two moles of hydrogen atoms to form 2 moles of water.
So, the molar ratio of oxygen to hydrogen is (1 to 2).
The mass of 1 mole of oxygen = (no. of moles)(molar mass) = (1 mol)(32.0 g/mol) = 32.0 g.The mass of 2 moles of hydrogen = (no. of moles)(molar mass) = (2 mol)(2.0 g/mol) = 4.0 g.
So, the mass ratio of oxygen to hydrogen (32.0 g/4.0 g) = (8: 1).
The barium isotope 133Ba has a half-life of 10.5 years. A sample begins with 1.1×1010 133Ba atoms. How many are left after (a) 6 years, (b) 10 years, and (c) 200 years?
Answer:
(a) 7.4 x 10⁹ atoms.
(b)
(c)
Explanation:
It is known that the decay of a radioactive isotope isotope obeys first order kinetics.Half-life time is the time needed for the reactants to be in its half concentration.If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).Also, it is clear that in first order decay the half-life time is independent of the initial concentration.The half-life of 133-Ba = 10.5 years.For, first order reactions:
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(10.5 years) = 0.066 year⁻¹.
Also, we have the integral law of first order reaction:kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 0.066 year⁻¹).
t is the time of the reaction.
[A₀] is the initial concentration of (133-Ba) ([A₀] = 1.1 x 10¹⁰ atoms).
[A] is the remaining concentration of (133-Ba) ([A] = ??? g).
(a) 6 years:
t = 6.0 years.
∵ kt = ln([A₀]/[A])
∴ (0.066 year⁻¹)(6.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])
∴ 0.396 = ln((1.1 x 10¹⁰ atoms)/[A]).
Taking exponential for both sides:
∴ 1.486 = ((1.1 x 10¹⁰ atoms)/[A]).
∴ [A] = (1.1 x 10¹⁰ atoms)/(1.486) = 7.4 x 10⁹ atoms.
(b) 10 years
t = 10.0 years.:
∵ kt = ln([A₀]/[A])
∴ (0.066 year⁻¹)(10.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])
∴ 0.66 = ln((1.1 x 10¹⁰ atoms)/[A]).
Taking exponential for both sides:
∴ 1.935 = ((1.1 x 10¹⁰ atoms)/[A]).
∴ [A] = (1.1 x 10¹⁰ atoms)/(1.935) = 5.685 x 10⁹ atoms.
(c) 200 years:
t = 200.0 years.
∵ kt = ln([A₀]/[A])
∴ (0.066 year⁻¹)(200.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])
∴ 13.2 = ln((1.1 x 10¹⁰ atoms)/[A]).
Taking exponential for both sides:
∴ 5.4 x 10⁵ = ((1.1 x 10¹⁰ atoms)/[A]).
∴ [A] = (1.1 x 10¹⁰ atoms)/(5.4 x 10⁵) = 2.035 x 10⁴ atoms.
Final answer:
To determine the number of 133Ba atoms left after a certain time, the half-life formula is used. After 6 years, about 7.78×109 atoms remain; after 10 years, approximately 5.77×109 atoms; and after 200 years, just 2.09×104 atoms remain, showing the exponential decrease due to radioactive decay.
Explanation:
The barium isotope 133Ba has a half-life of 10.5 years. To determine how many atoms are left after a certain time period, we can use the half-life formula. The formula is:
N(t) = N0 * (1/2)^(t/T)Where N(t) is the number of atoms remaining after time t, N0 is the initial number of atoms, and T is the half-life of the isotope.
After 6 years: N(6) = 1.1×1010 * (1/2)^(6/10.5) = 1.1×1010 * (1/2)^(0.5714) = 7.78×109 atomsAfter 10 years: N(10) = 1.1×1010 * (1/2)^(10/10.5) = 1.1×1010 * (1/2)^(0.9524) = 5.77×109 atomsAfter 200 years: N(200) = 1.1×1010 * (1/2)^(200/10.5) = 1.1×1010 * (1/2)^(19.0476) = 1.1×1010 * (1/2)19 = 2.09×104 atomsWe can see that after 200 years, a negligible amount of the original 133Ba atoms remain due to the nature of radioactive decay.
The combustion of ethane (C2H6)(C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2CO2 are produced when 5.95 mol5.95 mol of ethane is burned in an excess of oxygen? moles of CO2:CO2:
Answer:
11.9 moles of carbon-dioxide will produced.
Explanation:
Given moles of ethane = 5.95 mol
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]
According to reaction 2 moles of ethane produces 4 moles of carbon-dioxide.
Then, 5.95 moles of ethane will produce:
[tex]\frac{4}{2}\times 5.95 mol=11.9 mol[/tex] of carbon-dioxide
11.9 moles of carbon-dioxide will produced.
First, the stoichiometric relationship in the balanced chemical equation between ethane (C2H6) and carbon dioxide (CO2) is identified. From this it is shown that 1 mole of ethane produces 2 moles of carbon dioxide. Therefore, 5.95 moles of ethane will produce 11.9 moles of CO2.
Explanation:The combustion of ethane, C2H6, is described by the balanced chemical equation: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g). This indicates a stoichiometric relationship that 2 moles of ethane produce 4 moles of carbon dioxide, or 1 mole of ethane produces 2 moles of carbon dioxide. Therefore, if 5.95 moles of ethane are combusted in excess oxygen, it would produce twice this amount in moles of carbon dioxide. Hence, the reaction would produce 11.9 moles of CO2.
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The elementary reaction 2H20(g)<--->2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2, and O2 reach 0.0700 atm, 0.00200 atm, and 0.00600 atm respectively. What is the value of the equilibrium constant at this temperature?
Answer:
4.9 x 10⁻⁶.
Explanation:
For the reaction:2H₂O(g) ⇄ 2H₂(g) + O₂(g),
'
P of H₂ = 0.002 atm, P of O₂ = 0.006 atm, and P of H₂O = 0.07 atm.
∴ The equilibrium constant (Kp) = (P of H₂)²(P of O₂)/(P of H₂O)² = (0.002 atm)²(0.006 atm)/(0.07 atm)² = 4.9 x 10⁻⁶.
Consider a buffer solution containing CH3COOH and CH3COO-, with an equilibrium represented by: CH3COOH(aq) + H2O (l) ←----→ H3O+ (aq) + CH3COO- (aq) Describe what occurs if a strong acid such as HNO3 is added to the system, including an explanation of the direction of the equilibrium shift. Describe what occurs if a strong base such as KOH is added, including an explanation of the direction of the equilibrium shift.
Answer:
Here's what I get.
Explanation:
(a) The buffer equilibrium
The equation for the buffer equilibrium is
[tex]\rm CH_{3}COOH(aq) + H$_{2}$O(l) $\, \rightleftharpoons \,$ CH$_{3}$COO$^{-}$(aq) + H$_{3}$O$^{+}$(aq)[/tex]
(b) Addition of acid
If you add a strong acid like HNO₃, you are increasing the concentration of hydronium ion.
Per Le Châtelier's Principle, the system will respond in such a way as to decrease the concentration of hydronium ion.
The position of equilibrium will shift to the left.
(c) Addition of base.
If you add a strong base like KOH, The hydroxide ions will react with the hydronium ions to form water.
The concentration of hydronium ions will decrease.
Per Le Châtelier's Principle, the system will respond in such a way as to increase the concentration of hydronium ions.
The position of equilibrium will shift to the right.
When a strong acid is added to a buffer solution containing CH3COOH and CH3COO-, the equilibrium shifts to the left. When a strong base is added, the equilibrium shifts to the right.
Explanation:A buffer solution containing CH3COOH and CH3COO- can resist changes in pH when small amounts of a strong acid or base are added. When a strong acid such as HNO3 is added to the buffer system, the equilibrium will shift to the left, favoring the formation of more CH3COOH to consume the added H3O+ ions. This maintains the pH of the buffer solution. On the other hand, when a strong base such as KOH is added to the buffer system, the equilibrium will shift to the right, favoring the formation of more CH3COO- ions to consume the added OH- ions. This also helps to maintain the pH of the buffer solution.
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You encounter a solution that is acidic and you decide to test it by adding a small amount of a strong acid. The pH lowers slightly but is approximately unchanged, and still remains acidic. What can you say about the solution? a. It is a buffer solution. b. It is not a buffer solution It is a strong acid solution. d. The solution has been neutralized. e. The solution has excess acid present
Answer:
The solution on this question must be a buffer solution.
Explanation
Buffer solution has a molecule that able to withstand pH changes when a small amount of strong acid is added. In the solution, there is conjugate base that could react with H+. When a strong acid is added, the reaction shifts to the left so the conjugate eats up a bit H+ resulting in a lower amount of expected pH changes.
For the gas phase decomposition of nitrosyl chloride at 400 K , the rate of the reaction is determined by measuring the appearance of Cl2 . 2 NOCl= 2 NO + Cl2 At the beginning of the reaction, the concentration of Cl2 is 0 M. After 1.64×103 min the concentration has increased to 3.39×10-3 M. What is the rate of the reaction?
Answer:
2.067 x 10⁻⁷ M/min.
Explanation:
Knowing that the rate of the reaction is the change in the concentration of reactants (decrease) or the products (increase) with time.For the reaction: 2NOCl ⇄ 2NO + Cl₂,Rate of the reaction = - 1/2 d[NOCl]/dt = 1/2 d[NO]/dt = d[Cl₂]/dt
∵ d[Cl₂] = 3.39 x 10⁻³ M, dt = 1.64 x 10³ min.
∴ Rate of the reaction = d[Cl₂]/dt = (3.39 x 10⁻³ M) / (1.64 x 10³ min) = 2.067 x 10⁻⁷ M/min.
The rate of the reaction for the decomposition of nitrosyl chloride to nitrogen monoxide and chlorine is calculated by dividing the change in concentration of Cl₂ by the change in time, which yields a rate of 2.07×10⁻¶ M/min.
The balanced chemical equation is 2 NOCl → 2 NO + Cl₂. The reaction rate can be calculated using the change in concentration of Cl₂ over the change in time, as the rate of appearance of Cl₂ is directly related to the rate of the reaction. Since Cl₂ starts at 0 M and increases to 3.39×10⁻³ M over a period of 1.64×10³ minutes, you can calculate the average rate as follows:
Rate = Δ[Cl₂] / Δtime = (3.39×10⁻³ M - 0 M) / (1.64×10³ min) = 2.07×10⁻¶ M/min.
When 10.0 g of NH3 reacts, the actual yield of N2 is 8.50 g. What is the percent yield? 4 NH3 (g) + 6 NO(g) --> 5 N2 (g) + 6 H2O(l)
Answer:
%Yield = 41.3% (w/w)
Explanation:
4NH₃ + 6NO => 5N₂ + 6H₂O
10g NH₃ = (10/17)mole NH₃ = 0.588 mole NH₃ => 5/4(0.588)mole N₂ = 0.735 mole N₂ x 28g N₂/mole N₂ = 20.58g N₂ (Theoretical Yield)
Given Actual Yield = 8.5g N₂
%Yield = (Actual Yield/Theoretical Yield)100% = (8.5g N₂/20.58g N₂)100% =41.3% Yield (w/w)
The percent yield of N₂ is 41.34%
From the question,
We are to determine the percent yield of N₂
The given balanced chemical equation for the reaction is
4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)
This means 4 moles of NH₃ reacts with 6 moles of NO to produce 5 moles of N₂ and 6 moles of H₂O
First, we will determine the number of moles of NH₃ that reacted
From the question,
Mass of NH₃ that reacted = 10.0 g
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of NH₃ = 17.031 g/mol
∴ Number of moles of NH₃ that reacted = [tex]\frac{10.0}{17.031 }[/tex]
Number of moles of NH₃ that reacted = 0.58716 mole
Now, we will determine the theoretical number of moles of N₂ produced
From the balanced chemical equation
Since
4 moles of NH₃ reacts to produce 5 moles of N₂
Then,
0.58716 mole of NH₃ will react to produce [tex]\frac{0.58716 \times 5}{4}[/tex] moles of N₂
[tex]\frac{0.58716 \times 5}{4} = 0.73395[/tex]
∴ The theoretical number of moles of N₂ that would be produced is 0.73395 mole
Now, we will determine the theoretical yield in grams
Using the formula
Mass = Number of moles × Molar mass
Molar mass of N₂ = 28.0134 g/mol
∴ Mass of N₂ that would be produced = 0.73395 × 28.0134
Mass of N₂ that would be produced = 20.56 g
∴ The theoretical yield of N₂ is 20.56 g
Now, for the percent yield
[tex]Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield}\times 100\%[/tex]
From the question,
Actual yield of N₂ = 8.50 g
∴ Percent yield of N₂ = [tex]\frac{8.50}{20.56}\times 100\%[/tex]
Percent yield of N₂ = 41.34 %
Hence, the percent yield of N₂ is 41.34%
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A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. Physical states are optional. chemical equation: What is the limiting reactant? calcium hydroxide hydrochloric acid How many grams of salt are formed after the reaction is complete? mass of salt: gg How many grams of the excess reactant remain after the reaction is complete? excess reactant remaining:
Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.
Explanation:
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For calcium hydroxide:
Given mass of calcium hydroxide = 9.27 g
Molar mass of calcium hydroxide = 74.093 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol[/tex]
To calculate the moles of a solute, we use the equation:[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Volume of hydrochloric acid = 38.5mL = 0.0385 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 0.500 moles/ L
Putting values in above equation, we get:
[tex]0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol[/tex]
For the given chemical equation:[tex]2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)[/tex]
Here, the solid salt is calcium chloride.
By Stoichiometry of the reaction:
2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.
So, 0.01925 moles of hydrochloric acid will react with = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] of calcium hydroxide.
As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.
Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 molesBy Stoichiometry of the reaction:
2 moles of hydrochloric acid produces 1 mole of calcium chloride.
So, 0.01925 moles of hydrochloric acid will produce = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] of calcium chloride.
Now, calculating the mass of calcium chloride from equation 1, we get:
Molar mass of calcium chloride = 110.98 g/mol
Moles of calcium chloride = 0.009625 moles
Putting values in equation 1, we get:
[tex]0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g[/tex]
Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.
HCl is a limiting reactant
mass of salt: 1.068375 g
8.559 g of the excess reactant (Ca(OH)₂)remain after the reaction is complete
Further explanationThe reaction equation is the chemical formula of reagents and product substances
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products
Terms used:
Mole
The mole itself is the number of particles contained in a substance amounting to 6.02.10²³
[tex] \large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}} [/tex]
We determine the mole of each reactant to determine the limiting reactant
then:
mol Ca (OH₂ = mass: molar mass
mole of Ca (OH)₂ = 9.27 g: 74
mole of Ca (OH)₂ = 0.1253
mole HCl: 38.5 ml x 0.5 M = 19.25 mlmol = 0.01925 mol
From the number of moles, it can be seen that HCl is a limiting reactant
Reaction:
Ca(OH)₂ + 2HCl ⇒ CaCl₂ + 2H₂O
initial mole 0.1253 0.01925
reaction 0.009625 0.01925 0.009625 0.01925
remaining 0.115675 - 0.009625 0.01925
Remaining unreacted Ca (OH)₂ mole: 0.115675
Amount of mass of CaCl₂ salt formed:
CaCl₂ mass = mole x molar mass
CaCl₂ mass = 0.009625 x 111
CaCl₂ mass = 1.068375
Remaining Ca(OH)₂ = 0.115675 x 74 = 8.559 g
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A hydrogen atom has 1 electron. How many bonds can hydrogen form? A) 1 B) 2 C) 3 D) 4 E) 5
What is the standard molar enthalpy change of formation of O2(g) at 25 °C?
Answer:
Zero
Explanation:
Standard Heat of Formation (ΔH°f) for any element in it's basic standard state is zero (0) Kj/mole. If you have a college general chemistry text, look in the appendix for table with title 'Thermodynamic Properties of Substances at 25°C'. Scan down under the symbol (ΔH°f) ... Note that the substance listed by the 'zero' Kj/mole values will be an element. This is the element in basic standard state. Oxygen (O₂(g)) will have a 0 Kj/mole value as does H₂(g), N₂(g), F₂(g), Cl₂(g), Br₂(l) and I₂(s). These are the 7 diatomic elements in basic standard state. Look at the pattern that is made for this set of elements on the periodic table. Starting with N > O > F > Cl > Br > I it traces as a '7' => You just have to remember Hydrogen is a member of the set also.
An ecologist takes water samples from three different lakes in a mining area (A, B, and C). The hydrogen ion concentrations are 1 × 10–6 M, 1 × 10–5 M, and 1 × 10–4 M for lakes A, B, and C, respectively. What is the pH of Lake B?
Answer: The pH of Lake B is 5.
Explanation:
pH is defined as negative logarithm of hydrogen ion concentration. It is basically defined as the power of hydrogen ions in a solution.
Mathematically,
[tex]pH=-\log[H^+][/tex]
We are given:
Hydrogen ion concentration of Lake A, [tex][H_A]=1\times 10^{-6}M[/tex]
Hydrogen ion concentration of Lake B, [tex][H_B]=1\times 10^{-5}M[/tex]
Hydrogen ion concentration of Lake C, [tex][H_C]=1\times 10^{-4}M[/tex]
Putting values of hydrogen ion concentration for Lake B in above equation, we get:
[tex]pH=-\log(10^{-5})\\\\pH=5[/tex]
Hence, the pH of Lake B is 5.
You need to prepare an acetate buffer of pH 5.50 from a 0.872 M acetic acid solution and a 2.41 M KOH solution. If you have 580 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.50 ? The p????a of acetic acid is 4.76.
Answer:
Need to add 178ml of 2.41M KOH into the 580ml of 0.872M HOAc => Buffer Solution with pH = 5.50.
Explanation:
1. Determine the Base to Acid Ratio using the Henderson-Hasselbalch Equation
Given pH(Bfr) = 5.50; pKa(HOAc) = -logKa = -log(1.8x10ˉ⁵) = 4.76
pH(Bfr) = pKa(HOAc) + log([OAcˉ]/[HOAc]) => 5.5 = 4.76 + log([OAcˉ]/[HOAc])
=> log([OAcˉ]/[HOAc]) = 5.50 – 4.76 = 0.74 => [OAcˉ]/[HOAc] = 10^0.74 = 5.495*
*For an HOAc/OAcˉ Buffer to have a pH = 5.50 the [OAcˉ] concentration must be 5.495 times greater than the [HOAc] concentration.
2. Determine moles of KOH need be added to 580ml of 0.872M HOAc such that the moles of OAc⁻ is 5.495 times greater than moles of HOAc. The Acid/Base Rxn is HOAc + KOH => KOAc + H₂O.ˉ
That is, given 580ml(0.872M HOAc) + V(L)∙2.41M KOH => 5.495 x moles HOAc
=> 0.58(0.872)mole HOAc + X moles KOH => 5.495 x 0.58(0.872)mole KOAc (=OAcˉ)
=> 0.506 mole HOAc + X moles KOH* => 2.7805 mole KOAc ( = 2.7805 mole OAcˉ)
*note => KOH is the limiting reactant and will be consumed when added into HOAc solution leaving HOAc with OAcˉ that constitutes the buffer solution. The moles of KOH added is equal to the moles of OAcˉ produced. [HOAc] will decrease and [OAcˉ] will increase until the OAcˉ concentration is 5.495 times grater than the HOAc concentration.
3. After adding X moles of KOH, the following solution results …
=> (0.506 mole – X mole)HOAc + X mole OAcˉ
=> Since the moles OAcˉ is 5.495 x moles HOAc, the following linear expression in one unknown is generated …
=> moles OAcˉ produced/moles HOAc remaining = X / 0.506 – X = 5.495 where X = OAcˉ produced from rxn and 0.506 – X is the HOAc remaining. The moles of OAcˉ must be 5.495 times greater than moles of HOAc.
Solving for X => X = 5.495(0.506 – X) = 2.7805 – 5.495X => X = 2.7805/6.495 = 0.428 mole OAcˉ produced. The 0.428 mole OAcˉ is 5.495 time greater than (0.506 - 0.428) mole HOAc remaining.
4. This means that 0.428 mole KOH is needed for reaction with 580ml of 0.872M HOAc to give 0.428 mole OAcˉ with (0.506 – 0.482)mole HOAc remaining after mixing solution.
The volume of 2.41M KOH needed to deliver 0.428 mole KOH is V(KOH)∙2.41M = 0.428 mole => V(KOH) = (0.428/2.41)Liters = 0.178 Liter = 178ml of 2.41M KOH.
5. Verification of Results …
580ml(0.872M HOAc) + 178ml(2.41M KOH)
=> 0.58(0.872)mole HOAc + 0.178(2.41)mole KOH
=> 0.506 mole HOAc + 0.428 mole KOH
=> (0.506 – 0.428)mole HOAc + 0.428mole OAc⁻
=> 0.078 mole HOAc + 0.428mole OAcˉ
=> (0.078mol HOAc)/(0.58 + 0.178)Liter Soln + (0.428 mole OAcˉ)/(0.58 + 0.178)Liter Soln
=> (0.078mole/0.758L)HOAc + (0.428mole/0.872L)OAcˉ
=> 0.1029M HOAc + 0.5646M OAcˉ (Buffer Solution with pH = 5.50)
Checking pH of this buffer solution…
HOAc ⇄ H⁺ + OAcˉ
C(eq) 0.1029M [H⁺] 0.5646M
Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]
= [(1.8 x 10ˉ⁵)(0.1029)/(0.5646)]M = 3.28 x 10ˉ⁶M
pH = -log[H⁺] -log(3.28 x 10ˉ⁶) = 5.50
A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data. Zn(s)+2HBr(aq)⟶ZnBr2(aq)+H2(g)
Answer:
The standard enthalpy is -247KJ/mol.
Explanation:
The balanced equation of the reaction is :
Zn(s) + 2HBr(aq) ---> ZnBr2(aq) + H2(g)
Number of moles can be calculated by the formula:
[tex]No. of moles=\frac{{given mass}}{molecular mass}[/tex]
No. of moles of zinc = [tex]\frac{2.50}{65.88}[/tex]
No. of moles of zinc = 0.0382 moles.
No. of moles of HBr = [tex]\text{No. of mole}=molarity\times \text{volume of solution}[/tex]
No. of moles of HBr = 2.00×0.100
= 0.200
Here, the limiting reagent is Zinc because every mole of zinc used in the reaction twice of the moles of HBr are needed. HBr is present in high amounts as compared with Zinc.
Heat absorption can be calculated by:
Heat absorption= Total heat capacity × Temperature
= 448×21.1
=9452 J.
Standard enthalpy can be calculated by:
Standard enthalpy = [tex]\frac{Heat absorption}{No. of moles of Zn}[/tex]
=[tex]\frac{9452}{0.0382}[/tex]
=247.
The standard enthalpy of reaction is -247 KJ/mol as the heat has been given off in the reaction.
The standard enthalpy of the reaction using the data. Zn(s)+2HBr(aq)⟶ZnBr2(aq)+H2(g) is:
-247KJ/mol.Standard enthalpy is the rate of change of enthalpy with the emergence of one mole of the composites of its elements.
Furthermore, in order to balance the equation, we would have to combine the powdered zinc and the hydrobromic acid which will give us:
Zn(s) + 2HBr(aq) ---> ZnBr2(aq) + H2(g)Additionally, to calculate the number of moles, we would use the formula:
Number of moles= mass/molecular mass
Listing out the values of each property
Number of moles of zinc= 0.0382 moles
Number of moles of HBr= molarity * volume of solution
=0.200 moles
Heat absorption= Total heat capacity * Temperature
448 * 21.1
=9452 J
Therefore, to calculate the standard enthalpy, we would get:
Heat absorption/no of moles of zinc
This would give us 9452J/0.0382= 247 Kj/mol
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