In sodium chloride, the distance between the center of the sodium ion and the centerof an adjacent chloride ion is 2.819 angstroms. Calculate the density in g/cm3of an ideal NaCl crystal from this information and what you learned from this lab.Hints: To calculate mass, determine how many equivalent ions are in a unit cell. To determinevolume of the unit cell, start by determining the length of on side of the unit cell.
Answers
Answer:
[tex]\boxed{\text{2.17 g/cm}^{3}}[/tex]
Explanation:
1. Ions per unit cell
(a) Chloride
8 corners + 6 faces
[tex]\text{No. of Cl$^{-}$ ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}[/tex]
(b) Chloride
12 edges + 1 centre
[tex]\text{No. of Na$^{+}$ ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}[/tex]
There are four formula units of NaCl in a unit cell.
2. Mass of unit cell
m = 4 × NaCl = 4 × 58.44 u = 233.76 u
[tex]m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}[/tex]
3. Volume of unit cell
(a) Edge length
[tex]a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}[/tex]
(b) Volume
[tex]V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}[/tex]
4. Density
[tex]\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}[/tex]
Related Questions
) 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.
Answers
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
Learn more about mole fractions here:
https://brainly.com/question/31285244
#SPJ6
Calculate the number of pounds of CO2 released into the atmosphere when a 22.0 gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18, and that the density of gasoline is 0.692 g⋅mL^−1. This assumption ignores additives. Also, assume complete combustion. CO2 released:
Answers
Answer:
391.28771 pounds of carbon-dioxide released into the atmosphere.
Explanation:
Density of the gasoline ,d= 0.692 g/mL
Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL
Let mass of the gasolin be M
[tex]d=\frac{M}{V}[/tex]
M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g
Assume that gasoline is primarily octane (given)
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
Mass of octane burnt in the tank = M = 57,629.081 g
Moles of octane =[tex]\frac{57,629.081 g}{114.08 g/mol}=505.1637 mol[/tex]
According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.
Then 505.1637 mol of octane will give:
[tex]\frac{16}{2}\times 505.1637 mol=4,041.3100 mol[/tex] of carbon-dioxide
Mass of 4,041.3100 mol of carbon-dioxide:
4,041.3100 mol × 44.01 g/mol = 177,858.05 g
Mass of carbon-dioxide produced in pounds = 391.28771 pounds
391.28771 pounds of carbon-dioxide released into the atmosphe
The number of pounds of CO₂ released into the atmosphere when a 22.0 gallon tank of gasoline is burned in an automobile engine is 392.3 pounds.
We'll begin by converting 22 gallon to mL.
1 gallon = 3785.412 mL
Therefore,
22 gallon = 22 × 3785.412
22 gallon = 83279.064 mL
Next, we shall determine the mass of 83279.064 mL of gasoline (C₈H₁₈).
Density = 0.692 g/mL
Volume = 83279.064 mL
Mass of C₈H₁₈ =?Mass = Density × Volume
Mass of C₈H₁₈ = 0.692 × 83279.064
Mass of C₈H₁₈ = 57629.11 gNext, we shall determine the mass of C₈H₁₈ that reacted and the mass of CO₂ produced from the balanced equation.
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
Molar mass of C₈H₁₈ = (12×8) + (1×18)
= 96 + 18
= 114 g/mol
Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g
Molar mass of CO₂ = 12 + (16×2)
= 12 + 32
= 44 g/mol
Mass of CO₂ from the balanced equation = 16 × 44 = 704 g
Thus,
From the balanced equation above,
228 g of C₈H₁₈ reacted to produce 704 g of CO₂
Next, we shall determine the mass of CO₂ produced by the reaction of 57629.11 g of C₈H₁₈. This can be obtained as follow:
From the balanced equation above,
228 g of C₈H₁₈ reacted to produce 704 g of CO₂
Therefore,
57629.11 g of C₈H₁₈ will react to produce = [tex]\frac{57629.11 * 704}{228}[/tex] = 177942.515 g of CO₂
Finally, we shall convert 177942.515 g of CO₂ to pounds.
453.592 g = 1 pound
Therefore,
177942.515 g = [tex]\frac{177942.515}{453.592}[/tex]
177942.515 g = 392.3 pounds
Therefore, the number of pounds of CO₂ released into the atmosphere is 392.3 pounds.
Learn more: https://brainly.com/question/13476603
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
Answers
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.
Learn more about Elementary reactions here:https://brainly.com/question/31651448
#SPJ11
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.
Answers
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.
Real gas expansion in a vacuum is exothermic or endothermic?
first correct answer will be marked as brainliest!!!
plz respond ASAP!!!
Answers
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?
Answers
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
Learn more about equilibrium constant: https://brainly.com/question/12978582
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.
Answers
Answer:
ΔH(Combustion C₆H₆ = -3,262 Kj/mole
Explanation:
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
Answers
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.
4 NO + O2 + 2 H2O → 4 HNO2 The reaction above is second order in nitric oxide and first order in oxygen. How much faster will the rate be if the concentration of NO is tripled? 9 times faster 6 times faster 3 times faster 8 times faster
Answers
Answer:
Increases rate of rxn 9X
Explanation:
Given 4NO + O₂ + 2H₂O => 4HNO₂
Rate = k[NO]²[O₂] ; if [O₂] is kept constant => Rate = k[NO]²
For 2nd order reactions Rate = k[A]². If [A] = a, 2a, 3a … then the Rate trend for a 2nd order reaction is …
Rate 1 = k(a)² = ka²
Rate 2 = k(2a)² = 4ka²
Rate 3 = k(3a)² = 9ka² <= 3X increase in concentration
Rate 4 = k(4a)² = 16ka²
Increasing the concentration of NO by 3 times while keeping concentrations of O₂ and H₂O constant increases rate of reaction by 9 times.
The correct answer is that the rate will be 9 times faster if the concentration of NO is tripled.
The rate law for a reaction that is second order in nitric oxide (NO) and first order in oxygen (O2) can be written as:
[tex]\[ \text{Rate} = k[\text{NO}]^2[\text{O2}] \][/tex]
where[tex]\( k \)[/tex] is the rate constant, [tex]\([\text{NO}]\)[/tex] is the concentration of nitric oxide, and \[tex]([\text{O2}]\)[/tex] is the concentration of oxygen.
If the concentration of NO is tripled, the new rate can be calculated by substituting [tex]\( 3[\text{NO}] \)[/tex] into the rate law:
[tex]\[ \text{New Rate} = k(3[\text{NO}])^2[\text{O2}] \][/tex]
[tex]\[ \text{New Rate} = k \cdot 9[\text{NO}]^2 \cdot [\text{O2}] \][/tex]
[tex]\[ \text{New Rate} = 9 \cdot k[\text{NO}]^2[\text{O2}] \][/tex]
Since the original rate is [tex]\( k[\text{NO}]^2[\text{O2}] \),[/tex] the new rate is 9 times the original rate because[tex]\( 9 \cdot k[\text{NO}]^2[\text{O2}] \)[/tex]) is 9 times[tex]\( k[\text{NO}]^2[\text{O2}] \).[/tex]
Therefore, tripling the concentration of NO increases the rate of the reaction by a factor of 9. This is because the rate depends on the square of the concentration of NO, and when a concentration is tripled, its square becomes 9 times larger, resulting in a 9-fold increase in the rate of the reaction.
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)
Answers
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
Read more here:
https://brainly.com/question/13459509
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?
Answers
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.
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?
Answers
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.
A gas engine that operates on a Brayton cycle has an efficiency of 0.23. On a cold day, the temperature of the air drawn into the engine is 267 K.Part AWhat is the temperature of the air exhausted from the engine?
Answers
Answer:
347 KExplanation:
The Brayton cycle is Joule cycle and is the model for the operation of a gas turbine engine.
The general equation for the efficiency of an engine working between two temperatures is:
Efficiency = 1 - T cold / T hot.In this case:
T cold is the temperature at which the air is drawn from the atmosphere into the engine (turbine): 267 K; andT hot is the temperature of the air exhausted from the engine (the unknown variable of the problem).Thus, substituting the data into the formula, you get:
0.23 = 1 - 267 K / T hot267 K / T hot = 1 - 0.23267 K / T hot = 0.77T hot = 267 K / 0.77T hot = 347 KAnswer: the temperature of the air exhausted from the engine is 347 K
Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [HCl]eq = 0.13 M, [HI]eq = 5.6 × 10-16 M, [Cl2]eq = 0.0019 M. 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s)
Answers
The equilibrium constant, Kc, of the given reaction is calculated by substitifying the equilibrium concentrations into the Kc expression. Considering the chemical equation 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), the Kc expression is [HCl]^2 / ([HI]^2 * [Cl2]). Substituting the given equilibrium concentrations in, Kc equates to 4.67 × 10^33.
Explanation:The equilibrium constant (Kc) for a chemical reaction is a number that expresses the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. In the equation given: 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), I2 (s) is a pure solid and according to the rules of equilibrium, pure solids and liquids are not included in the equilibrium expression. Therefore, the equilibrium constant expression does not include I2.
For this reaction, the Kc expression is: Kc = [HCl]^2 / ([HI]^2 * [Cl2])
Substituting the equilibrium concentrations into this expression gives: Kc = (0.13 M)^2 / ((5.6 × 10^-16 M)^2 * 0.0019 M) = 4.67 × 10^33
Learn more about Chemistry Equilibrium Constant here:
https://brainly.com/question/36150835
#SPJ3
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
Answers
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
#SPJ3
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?
Answers
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.
Learn more about Carbon-14 Dating here:
https://brainly.com/question/32047290
#SPJ3
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:
Answers
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
Learn more
The mass of one mole of raindrops
brainly.com/question/5233234
moles of NaOH
brainly.com/question/4283309
moles of water you can produce
brainly.com/question/1405182
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?
Answers
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.
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?
Answers
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.
Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel has a pinhole-sized leak, which of the following will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the gas mixture has effused?
Answers
Answer:
VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Explanation:
Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.
=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Review Graham's Law => Effusion Rate ∝ 1/√formula mass.
The option that will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the gas mixture has effused is;
A) Pressure He < Pressure Ne < Pressure Ar
The missing options are;
a) Pressure He < Pressure Ne < Pressure Ar
b) Pressure He < Pressure Ar < Pressure Ne
c) Pressure Ar < Pressure Ne < Pressure He
d) Pressure He = Pressure Ar = Pressure Ne
Now, to answer this question, let us first write the formula for graham's law of diffusion/effusion.
r ∝ 1/√M
Where;
r is rate of diffusion/effusion
M is molar mass
In effusion, there is usually a barrier with very small holes that prevents the gas from expanding very fast into a new volume. Thus, we can say that the heavier the gas, the lesser the effusion rate and the lighter the gas, the faster the effusion rate.
Now, the pressure of the heavier gas is usually higher than that of the lighter ones.
The molar mass of the given gases are;
Molar mass of He = 4 g/mol
Molar mass of Ne = 20.18 g/mol
Molar mass of Ar = 39.95 g/mol
From the effusion equation, we see that the rate of effusion is inversely proportional to the molar mass.
Thus, the higher the molar mass, the lesser the effusion rate and as such, the higher the pressure since it is a heavy gas.
Thus, Ar has the highest Molar mass and will have the highest pressure. Next is Neon(Ne), then Helium(He).
Thus, in conclusion;
Pressure He < Pressure Ne < Pressure Ar
Read more at; https://brainly.com/question/19589867
A hydrogen atom has 1 electron. How many bonds can hydrogen form? A) 1 B) 2 C) 3 D) 4 E) 5
Answers
There is only 1 possible bond
Hope this helps
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:
Answers
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.
Learn more about Combustion of Ethane here:https://brainly.com/question/26891476
#SPJ3
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)
Answers
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%
Learn more here: https://brainly.com/question/19655772
Which of the following is a result of glycolysis? production of CO2 conversion of glucose to two, three-carbon compounds a net loss of two ATPs per glucose molecule conversion of NADH to NAD+
Answers
Answer:
the 2nd part
Explanation:
could conversion is called cellular respiration
Glycolysis results in the conversion of one glucose molecule into two pyruvate molecules, a net gain of two ATP molecules, and the reduction of NAD+ to NADH. It does not result in the production of CO2 or the conversion of NADH back to NAD+.
Explanation:Glycolysis is a metabolic pathway occurring in the cytosol of cells where one glucose molecule is converted into two, three-carbon compounds, specifically, Pyruvate. This process is anaerobic, meaning it doesn't require oxygen. There's also a net production of two ATP molecules per glucose molecule, so the option indicating a net loss is incorrect. Furthermore, in the process of glycolysis, the compound NAD+ is actually reduced to NADH, not the other way around. CO2 production doesn't occur in glycolysis itself but in subsequent processes of cellular respiration namely the Krebs cycle and oxidative phosphorylation.
Learn more about Glycolysis here:https://brainly.com/question/14089531
#SPJ6
98. Silicon has three naturally occurring isotopes: Si-28 with mass 27.9769 amu and a natural abundance of 92.21%, Si-29 with mass 28.9765 amu and a natural abundance of 4.69%, and Si-30 with mass 29.9737 amu and a natural abundance of 3.10%. Calculate the atomic mass of silicon.
Answers
Answer:
Wt Avg Atomic Wt = 28.0577 amu
Explanation:
Atomic Mass = ∑Wt Avg contributions of isotopes of an element
Si-28 => 92.21% => 0.9221 decimal fraction => Wt Avg Si-28 = 0.9221(27.9769) = 25.7695 amu
Si-29 => 4.69% => 0.0469 decimal fraction => Wt Avg Si-29 = 0.0469(28.9765) = 1.3590 amu
Si-30 => 3.10% => 0.0310 decimal fraction => Wt Avg Si-30 = 0.0310(29.9737) = 0.9292
Wt Avg Atomic Wt = ∑Wt Avg Contributions = (25.7695 + 1.3590 + 0.9292)amu = 28.0577 amu
Answer : The average atomic mass of silicon is, 28.08 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
For isotope Si-28 :
Mass of isotope Si-28 = 27.9769 amu
Fractional abundance of isotope Si-28 = 92.21 % = 0.9221
For isotope Si-29 :
Mass of isotope Si-29 = 28.9765 amu
Fractional abundance of isotope Si-29 = 4.69 % = 0.0469
For isotope Si-30 :
Mass of isotope Si-30 = 29.9737 amu
Fractional abundance of isotope Si-30 = 3.10 % = 0.0310
Putting values in equation 1, we get:
[tex]\text{Average atomic mass }=[(27.9769\times 0.9221)+(28.9765\times 0.0469)+(29.9737\times 0.0310)][/tex]
[tex]\text{Average atomic mass }=28.08amu[/tex]
Thus, the average atomic mass of silicon is, 28.08 amu
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
Answers
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)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?
Answers
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.
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? %
Answers
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.
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?
Answers
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]
What is the standard molar enthalpy change of formation of O2(g) at 25 °C?
Answers
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.