One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.00 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T = 300 K; 1 L·atm = 101.3 J)

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

1. Double replacement but no reaction

2. Double replacement but no reaction

3. Combustion

4. Combustion

5. Single replacement

6. Double replacement

I didn't solve the ones that are repeated! Hope it helped;)

Answer:

1. Double displacement.

2. Combustion.

3. Simple displacement.

4.Double displacement.

Explanation:

Hello,

1. [tex]2HBr(aq)+Ba(OH)_2(aq) -->2H_2O(l)+BaBr_2(aq)[/tex]

In this case, it is about a double displacement reaction since all the cations (H and Ba) and anions (Br and OH) are exchanged.

2. [tex]C_2H_4(g)+3O_2(g)-->2CO_2(g)+2H_2O(l)[/tex]

In this case, it is about the combustion of ethene.

3. [tex]Cu(s)+FeCl_2(aq)-->Fe(s)+CuCl_2(aq)[/tex]

In this case, it is about a simple displacement reaction since the iron (II) cations become solid iron and on the contrary for copper.

4. [tex]Na_2S(aq)+2AgNO_3(aq)-->2NaNO_3(aq)+Ag_2S(s)[/tex]

Finally, it is about a double displacement chemical reaction since the sodium and silver cations are exchanged with the sulfide and nitrate anions.

Best regards.

Answer : The mass of dry [tex]NH_4Cl[/tex] needed are, 174.196 grams.

Explanation :

First we have to calculate the pOH of the solution.

[tex]pH+pOH=14[/tex]

[tex]pOH=14-pOH\\\\pOH=14-8.72\\\\pOH=5.28[/tex]

Now we have to calculate the [tex]pK_b[/tex].

[tex]pK_b=-\log K_b[/tex]

[tex]pK_b=-\log (1.8\times 10^{-5})[/tex]

[tex]pK_b=4.745[/tex]

Now we have to calculate the concentration of base, [tex]NH_4Cl[/tex].

Using Henderson Hesselbach equation :

[tex]pOH=pKb+\log \frac{[Salt]}{[Base]}[/tex]

[tex]pOH=pKb+\log \frac{[NH_4Cl]}{[NH_3]}[/tex]

Now put all the given values in this equation, we get :

[tex]5.28=4.745+\log \frac{[NH_4Cl]}{0.500}[/tex]

[tex][NH_4Cl]=1.714M[/tex]

Now we have to calculate the mass of [tex]NH_4Cl[/tex].

Formula used :

[tex]Molarity=\frac{\text{Moles of }NH_4Cl}{\text{Volume of solution}}[/tex]

[tex]Molarity=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl\times \text{Volume of solution}}[/tex]

Now put all the given values in this formula, we get:

[tex]1.714M=\frac{\text{Mass of }NH_4Cl}{53.49\times 1.90L}[/tex]

[tex]\text{Mass of }NH_4Cl=174.196g[/tex]

Therefore, the mass of dry [tex]NH_4Cl[/tex] needed are, 174.196 grams.

The pH has been the hydrogen ion concentration and pOH is the hydroxide ion concentration. The mass of ammonium chloride in the solution is 174.196 grams.

The Kb is the base dissociation constant for the compound.

The pH of the solution is 8.72, the pOH of the solution is given as:

[tex]\rm pH=14-pOH\\8.72=14-pOH\\pOH=5.28[/tex]

The pOH of the given solution is 5.28. The concentration of ammonium chloride salt in the solution from pOH can be given as:

[tex]\rm pOH=pKb\;+\;log\dfrac{salt}{acid}[/tex]

The pKb has been given as the logarithmic value of Kb. The concentration of ammonia is 0.5 M. Substituting the values for the concentration of ammonium chloride salt as:

[tex]\rm 5.28=log\;1.8\;\times\;10^{-5}\;+\;log\dfrac{NH_4Cl}{0.5}\\\\ 5.28=4.745\;+\;log\dfrac{NH_4Cl}{0.5}\\\\NH_4Cl=1.714\;M[/tex]

The concentration of ammonium chloride salt is 1.714 M. The volume of the solution is 1.90 L. The molar mass of the compound is 53.49 g/mol.

Substituting the values for the mass of ammonium chloride as:

[tex]\rm Molarity=\dfrac{mass}{molar\;mass\;\times\;volume} \\\\1.714\;M=\dfrac{mass}{53.49\;g/mol\;\times\;1.90\;L}\\\\ Mass=174.196\;g[/tex]

The mass of ammonium chloride added to the solution is 174.196 grams.

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Answer:

False

Explanation:

For different diameter pipes flow rate remains constant but velocity is different. (from continuity A1V1= A2V2). As area changes the velocity of fluid changes

Answer:

-238.54 kJ/mol.

Explanation:

C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ? ?? kJ/mol.

(1) C(graphite) + O₂(g) → CO₂(g),                         ΔHf₁° = -393.5 kJ/mol .

(2) H2(g) + ½ O₂(g) → H₂O(l),                               ΔHf₂° = -285.8 kJ/mol .

(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l),    ΔH₃° = -726.56 kJ/mol .

(1) C(graphite) + O₂(g) → CO₂(g),                         ΔHf₁° = -393.5 kJ/mol .

(2) 2H2(g) + O₂(g) → 2H₂O(l),                               ΔHf₂° = 2x(-285.8 kJ/mol ).

(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g),     ΔH₃° = 726.56 kJ/mol .

The standard enthalpy of formation of liquid methanol, CH₃OH(l) =  ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol ) + 2(-285.8 kJ/mol ) - (- 726.56 kJ/mol) = -238.54 kJ/mol.

The standard enthalpy change of formation of liquid methanol, CH3OH(l), is calculated by manipulating and summing the reactions provided, considering the rules of Hess's law. The resulting ΔH° of the methanol formation from its elements is +976.0 kJ/mol.

To calculate the standard enthalpy of formation ΔH° of liquid methanol, CH3OH(l), we will use the concept of Hess's law and the provided chemical equations. Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the pathway or steps taken to achieve the conversion.

The standard enthalpy formation reaction is defined as the formation of 1 mol of a compound from its elemental forms under standard state conditions. For methanol, the reaction would be: C(graphite) + 2H2(g) + 1/2O2 ⟶ CH3OH(l)

None of the provided reactions exactly match this equation. However, their combination, while keeping in mind the stoichiometry, gives the desired reaction. Note that as we know enthalpy is a state function, we could 'add' or 'subtract' reactions to get the desired one.

The reverse of the CO2 formation reaction is: CO2(g) ⟶ C(graphite) + O2, ΔH° = +393.5 kJ/mol

The reaction for the formation of 2 mol of H2O needs to be halved: 1/2H2(g) + 1/2O2 ⟶ 1H2O(l), ΔH° = -285.8 kJ/mol/2 = -142.9 kJ/mol

The combustion of methanol reaction needs to be reversed: CO2(g) + 2H2O(l) ⟶ CH3OH(l) + O2(g), ΔH° = +726.4 kJ/mol

Adding these revised equations together gives the desired equation for the formation of methanol from its elements and the ΔH formation is the sum of the ΔH of these revised reactions δH°(formation of CH3OH) = +393.5 kJ/mol - 142.9 kJ/mol + 726.4 kJ/mol = +976. Omega kJ/mol.

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Answer:

Explanation:

For reaction stoichiometry problems like this, convert given data to moles and set up a ratio expression that relates to the balanced equation. That is ...

Given Rxn =>      Mg(OH)₂ + 2HCl => 2H₂O + MgCl₂

Given mass =>      3.26g = (3.26g)/(58g/mol) = 0.056 mole Mg(OH)₂

If from equation  1 mole Mg(OH)₂ reacts with 2 moles HCl

then,            0.056 mole Mg(OH)₂ reacts with x moles HCl

Setting up ratio and proportion expression ...

=> (1 mole Mg(OH)₂) /(0.056 mole Mg(OH)₂) = (2 moles HCl)/x

=> x = [2(0.056)/(1)] mole HCl neutralized = 0.112mole HCl (36g/mol)

= 4.05 grams of HCl neutralized.

4.08 grams of HCl are neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)₂.

Let's consider the neutralization reaction between HCl and Mg(OH)₂.

Mg(OH)₂(aq) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

First, we will convert 3.26 g of Mg(OH)₂ to moles using its molar mass (58.32 g/mol).

[tex]3.26 g \times \frac{1mol}{58.32g} = 0.0559 mol[/tex]

The molar ratio of Mg(OH)₂ to HCl is 1:2. The moles of HCl that react with 0.0559 moles of Mg(OH)₂ are:

[tex]0.0559 mol Mg(OH)_2 \times \frac{2molHCl}{1mol Mg(OH)_2} = 0.112molHCl[/tex]

Finally, we will convert 0.112 moles of HCl to grams using its molar mass (36.46 g/mol).

[tex]0.112 mol \times \frac{36.46g}{mol} = 4.08 g[/tex]

4.08 grams of HCl are neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)₂.

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Final answer:

To create a 90% confidence interval for the mean amount of mercury in tuna sushi, use the sample mean (0.836 ppm), sample standard deviation (0.253 ppm), and sample size to calculate the margin of error and apply it to the sample mean.

Explanation:

To construct a 90% confidence interval estimate of the mean amount of mercury in the population, you can use the sample mean (0.836 ppm) and the sample standard deviation (0.253 ppm) along with the sample size to calculate the margin of error. Then add and subtract this margin of error from the sample mean to find the confidence interval.

The formula for the confidence interval is:

Confidence Interval = Sample Mean ± (t * (Sample Standard Deviation / sqrt(Sample Size)))

You will need a t-score for the appropriate degree of freedom and desired confidence level, which you can find using statistical software or a t-distribution table. Once you have the t-score, you can calculate the margin of error and the confidence interval.

This method reflects the uncertainty inherent in estimating a population parameter from a sample statistic and assumes that the sample is representative of the population and that mercury levels are normally distributed.

Answer: The excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 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)

Given mass of sodium sulfide = 12.026 g

Molar mass of sodium sulfide = 78.045 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of sodium sulfide}=\frac{12.026g}{78.045g/mol}=0.154mol[/tex]

Given mass of mercury (II) perchlorate = 80.701 g

Molar mass of mercury (II) perchlorate = 399.49 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of mercury (II) perchlorate}=\frac{80.701g}{399.49g/mol}=0.202mol[/tex]

For the given chemical equation:

[tex]Hg(ClO_4)_2(aq.)+Na_2S(aq.)\rightarrow HgS(s)+2NaClO_4(aq.)[/tex]

Here, the solid precipitate is mercury sulfide.

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury (II) perchlorate.

So, 0.154 moles of sodium sulfide will react with = [tex]\frac{1}{1}\times 0.154=0.154moles[/tex] of mercury (II) perchlorate

As, given amount of mercury (II) perchlorate is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury sulfide.

So, 0.154 moles of sodium sulfide will react with = [tex]\frac{1}{1}\times 0.154=0.154moles[/tex] of mercury sulfide.

Now, calculating the mass of mercury sulfide from equation 1, we get:

Molar mass of mercury sulfide = 232.66 g/mol

Moles of mercury sulfide = 0.154 moles

Putting values in equation 1, we get:

[tex]0.154mol=\frac{\text{Mass of mercury sulfide}}{232.66g/mol}\\\\\text{Mass of mercury sulfide}=35.82g[/tex]

Hence, the excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.

We need to find the limiting reactant based on the moles of each reactant in the solution, based on which we can calculate the mass of the solid precipitate that will form and the mass of the reactant in excess that will remain.

To solve this problem, we bring in the concept of limiting reactants in a chemical reaction. The limiting reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product that can be formed. The reaction between mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfide (Na2S) results in the formation of a precipitate. Based on the provided masses, we first need to calculate the moles of each reactant. We'd then compare the stoichiometric ratios to determine the limiting reactant.

After determining the limiting reactant, we can use stoichiometry to calculate how much of the precipitate will form. Furthermore, we can calculate the remaining quantity of the reactant in excess after the reaction.

Please note that the exact calculations require the molecular weights of the reactants and knowledge about the balanced chemical equation of the reaction.

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Answer:

P₂O₅ + 3H₂O => 2H₃PO₄

Explanation:

Nothing to explain.

Mixed molarity = 0.25 M

Osmotic pressure solution: 6.15 atm

Osmotic pressure is the minimum pressure given to the solution so that there is no osmotic displacement from a more dilute solution to a more concentrated solution.

General formula:

[tex]\large{\boxed {\bold {\pi \: = \: M \: x \: R \: x \: T}}}[/tex]

π = osmotic pressure (atm)

M = solution concentration (mol / l)

R = constant = 0.08205 L atm mol-1 K-1

T = Temperature (Kelvin)

Mixing solution

To find the molarity of the mixed solution we can use the following formula:

Vc. Mc = V1.M1 + V2.M2

where

Mc =Mixed molarity

Vc = mixed volume

Mr. NaCl = 58.5

Mr. Dextrose = C₆H₁₂O₆ = 180

mole NaCl = gram / Mr

mole NaCl = 1.75 / 58.5 = 0.03

Dextrose mole = 40/180 = 0.22

Assuming 1 liter of solution

M mixture = mole NaCl + mole dextrose / 1 liter

M mixture = 0.03 + 0.22 / 1 L

M mix = 0.25 M

T = 25 + 273 = 298 K

The osmosis pressure of the solution becomes:

π = 0.25. 0.0825. 298 = 6.15 atm

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Grade: Senior High School

Subject: Chemistry

Chapter: Colligative property

Keywords: osmotic pressure, molarity, mole,dextrose, NaCl

Answer: The volume of CO formed is 254.43 L.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

Given mass of [tex]Ni(CO)_4[/tex] = 444 g

Molar mass of [tex]Ni(CO)_4[/tex] = 170.73 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol[/tex]

For the given chemical reaction:

[tex]Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)[/tex]

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = [tex]\frac{4}{1}\times 2.60=10.4mol[/tex] of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = [tex]62.364\text{ L Torr }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = [tex]22^oC=(273+22)K=295K[/tex]

Putting values in above equation, we get:

[tex]752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L[/tex]

Hence, the volume of CO formed is 254.43 L.

The volume of CO gas formed from the complete decomposition of 444 g of Ni(CO)4 can be found using the ideal gas law.

The given equation shows the thermal decomposition of nickel tetracarbonyl, Ni(CO)4: Ni(CO)4 (l) → Ni (s) + 4CO (g)

To find the volume of CO gas formed, we need to use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

We can use the given information to find the number of moles of CO, and then use the ideal gas law to find the volume. First, we need to convert the mass of Ni(CO)4 to moles using the molar mass:

Moles of Ni(CO)4 = mass of Ni(CO)4 / molar mass of Ni(CO)4

Next, we can use the stoichiometry of the balanced equation to relate the moles of Ni(CO)4 to moles of CO:

Moles of CO = moles of Ni(CO)4 × (4 moles of CO / 1 mole of Ni(CO)4)

Finally, we can use the ideal gas law to find the volume of CO:

V = (n × R × T) / P

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Answer:

For A: Aluminium is the limiting reagent.

For B: Oxygen gas is the limiting reagent.

For C: Aluminium is the limiting reagent.

For D: Aluminium is the limiting reagent.

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 1 mole of aluminum will react with = [tex]\frac{3}{4}\times 1=0.75moles[/tex] of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

By stoichiometry of the reaction:

3 moles of oxygen gas reacts with 4 moles of aluminium

So, 2.6 moles of oxygen gas will react with = [tex]\frac{4}{3}\times 2.6=3.458moles[/tex] of aluminium.

As, the given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

Hence, oxygen gas is the limiting reagent.

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 16 mole of aluminum will react with = [tex]\frac{3}{4}\times 16=12moles[/tex] of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 7.4 mole of aluminum will react with = [tex]\frac{3}{4}\times 7.4=5.55moles[/tex] of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

Limiting reactant are defined as the reactant that is involved primarily in the reaction and formation of product depends on this reactant. the limiting reactant for the given equation are as follows:

Limiting reagent are those that limits the formation of products, whereas the reagent present in the excess amount are known as excess reagent.

For A:

For the given chemical reactions, limiting reagents are:

By applying the stereochemistry concept:

4 moles of Al reacts with 3 moles of oxygen.

Hence, the given amount of oxygen is more than the required amount, it is an excess reagent and aluminum is limiting reagent.

Similarly, in the chemical reaction, where 4 moles of aluminum react with 2.6 moles of oxygen, such that:

3 moles of oxygen reacts with 4 moles of aluminum.

The given amount of aluminium is more than the required amount, aluminum is considered as an excess reagent and oxygen gas is the limiting reagent.

Also, 16 moles of aluminum reacts with 13 moles of oxygen.

4 moles of aluminum reacts with 3 moles of oxygen, such that:

Thus, oxygen is excess reagent and aluminum is limiting reagent.

7.4 mol Aluminum reacts with 6.5 moles of oxygen, such that:

Therefore, the aluminum is limiting reagent and oxygen is excess reagent.

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Answer:

[tex]r = k [A]^{2}[B]^{2}[/tex]

Explanation:

A + B + C ⟶ D

[tex]\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}[/tex]

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

[tex]\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}[/tex]

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

[tex]\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}[/tex]

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

[tex]\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}[/tex]

The rate law for the given reaction is determined to be rate = k[A]^2[B]^2[C]^0, based on careful comparisons of the relationship between initial concentrations of reactants [A], [B], [C] and the observed initial reaction rates in different experiments. The reaction is found to be second order in [A] and [B], and zero order in [C].

The rate law for the given chemical reaction can be derived by comparing the changes in the experimentally determined initial rates with changes in the initial concentrations of the reactants [A], [B], and [C]. It's important to note here that the rate law shows how the rate of a reaction depends on the concentration of its reactants.

Looking at Experiments 1 and 2, we observe that the initial concentration of [A] has doubled (from 0.0500 mol/L to 0.100 mol/L), while the initial concentrations of [B] and [C] remain constant. The initial reaction rate quadruples, which suggests that the reaction is second order with respect to [A].

Comparing Experiments 2 and 3, the initial concentration of [B] doubles (from 0.0500 mol/L to 0.100 mol/L), while the concentrations of [A] and [C] remain constant. The rate quadruples, which shows that the reaction order is also second in [B].

Now comparing Experiments 1 and 4, where [C] doubles and [A] and [B] are kept constant, the rate remains unchanged. This suggests the reaction order is zero in [C].

Adding these together, we conclude that the rate law is rate = k[A]^2[B]^2[C]^0, where k is the rate constant.

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The reason why the entropy of a system increases when it is compressed or when a liquid evaporates is because these processes increase the disorder of the system.

When a gas is compressed, the molecules are forced closer together, which means that there are more ways for them to be arranged. Similarly, when a liquid evaporates, the molecules are released from the liquid state and enter the gas state, which means that there are even more ways for them to be arranged.

The reason why the entropy of a system may decrease when it is cooled and expanded is because these processes can decrease the disorder of the system. When a system is cooled, the molecules move more slowly, which means that there are fewer ways for them to be arranged. Similarly, when a system is expanded, the molecules have more space to move around, which also means that there are fewer ways for them to be arranged.

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Answer:

[tex]1.9 \times 10^{6}\text{ m/s}[/tex]

Explanation:

[tex]v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}[/tex]

Data:

T = 3.02 × 10⁸ K

M = 2.013 × 10⁻³ kg/mol

Calculation:

[tex]v_{\text{rms}} = \sqrt{\dfrac{3\times 8.314\text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times 3.02 \times 10^{8} \text{ K}}{2.014 \times 10^{-3} \text{ kg}\cdot \text{mol}^{-1}}}\\\\\\=\sqrt{3.740\times 10^{12} \text{ (m/s)}^{2}} = 1.9 \times 10^{6}\text{ m/s}[/tex]

The thermal energy and the conservation of energy allows to find the average speed of the deuteron atoms is:

            v = 1.9 10⁶ m / s

The thermal energy of a particle is given by the Boltzmann energy partition relation, which in three dimensions is:

           E = [tex]\frac{3}{2}[/tex]  kT

Energy kinetic s the energy of movement and its expression is:

           K = ½ m v²

They indicate that the temperature of the plasma is T = 3.02 10⁸ K.

If there are no losses, the energy is conserved.

          K = E

          ½ m v² = [tex]\frac{3}{2}[/tex]  kT

          v = [tex]\sqrt{\frac{3 kT}{m} }[/tex]  

[tex]\frac{1.66 \ 10^{-27} kg}{1 u}[/tex]

The mass of deuteron is m = 2.013 u

Let's reduce to kg

            m = 2.013 u (  [tex]\frac{1.66 \ 10^{-27} kg}{1 uy}[/tex])

            m = 3.5358 10⁻²⁷ kg

We take the mass of a deuterium to 1 mole, multiplying by Avogador's number.

             m = 3.5358 10⁻²⁷  6.022 10²³

             m = 2.129 10⁻³ kg / mol

We calculate

             v = [tex]\sqrt{ \frac{3 \ 8.314 \ 3.02 \ 10^8 }{2.129 \ 10^{-3} } }[/tex]  

             v = [tex]\sqrt{3.538 \ 10^{12}}[/tex]

             v = 1.88 10⁶ m / s

They ask for the result with two significant figures.

             v = 1.9 10⁶ m / s

In conclusion using thermal energy and conservation of energy we can find the average speed of deuteron atoms is:

            v = 1.9 10⁶ m / s

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Answer : The value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

Explanation : Given,

Moles of gas = 10 mole

Initial pressure of gas = 1 atm

Final pressure of the gas = 10 atm

Temperature of the gas = [tex]0^oC=273+0=273K[/tex]

According to the question, this is the case of isothermal reversible compression of gas.

As per first law of thermodynamic,

[tex]\Delta U=q+w[/tex]

where,

[tex]\Delta U[/tex] = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

[tex]w=-2.303nRT\log (\frac{P_1}{P_2})[/tex]

where,

w = work done on the gas

n = number of moles of gas

R = gas constant = 8.314 J/mole K

T = temperature of gas

n = moles of the gas

[tex]P_1[/tex] = initial pressure of gas

[tex]P_2[/tex] = final pressure of gas

Now put all the given values in the above formula, we get the work done.

[tex]w=-2.303\times 10mole\times 8.314J/moleK\times 273K\times \log (\frac{1atm}{10atm})[/tex]

[tex]w=52271.69J[/tex]

And we know that, the heat is equal to the work done with opposite sign convention.

So, [tex]q=-52271.69J[/tex]

Therefore, the value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

Answer:

Wt%H₂SO₄ = 10.2% (w/w)

Explanation:

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent = 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

Wt%H₂SO₄ = (113.68 g/1113.68g)100% = 10.2% (w/w)

The percentage of sulfuric acid = 10.2% (w/w)

Given:

Molality of solution = 1.61m

It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent

= 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution is made up of solute and solvent. Thus,

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

[tex]Wt\% \text{ of } H_2SO_4 = \frac{113.68 g}{1113.68g}*100\% = 10.2\% (w/w)[/tex]

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Answer:

The answer is 357.85 K.

Explanation:

In order to convert temperatures from degrees Celsius to degrees Kelvin, use the following formula:

temperature in degrees Kelvin = temperature in degrees Celsius + 273.15.

(Oftentimes, 273 is used instead of 273.15; it depends on the discretion of the teacher or student.)

The temperature given in degrees Celsius = 84.7° C

Therefore, the temperature in degrees Kelvin (K) = 84.7 + 273.15 = 357.85 K.

Answer:

A. One or two letters are used to represent the name of each element

APEX

One or two letters are used to represent the name of each

element

The name of each element in the periodic table is symbolized by one or two letters.

There are different ways of achieving this. Sometimes;

These are some of the ways of symbolizing the name of each element.

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Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

Answer:

Helium

Explanation:

Final answer:

To determine which reactant is in excess, compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation. In this case, NO2 is in excess.

Explanation:

To determine which reactant is in excess, we need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.

The balanced equation is:

3NO2(g) + H2O(l) -> 2HNO3(l) + NO(g)

According to the equation, the ratio of NO2 to H2O is 3:1. So, for every 3 moles of NO2, we need 1 mole of H2O.

Given that there are 4.2 mol of NO2 and 0.50 mol of H2O, we can calculate the mole ratio of NO2 to H2O:

4.2 mol NO2 / 3 mol NO2 = 1.4 mol NO2

0.50 mol H2O / 1 mol H2O = 0.50 mol H2O

From the calculations, it is clear that the amount of NO2 is more than the required amount based on the stoichiometric ratio. Therefore, NO2 is in excess.

Answer:

C

Explanation:

Hydronium (H₃O+) is the same as H+ (aq) because of the same net charge. The acidic property of a solvent or solution is governed the amount of H+ ion in it. The higher the H+ ions the lower the pH. In pure water H₃O+ and OH- are in equal amount. More of OH- ions turn the solution basic.

Answer:

ΔH = 57 Kj/mole H₂O

Explanation:

Answer:

ΔH/mol H₂O = 55346 J/mol H₂O =55.346 kJ/mol H₂O

Explanation:

The reaction that occurs in this case is:

Ba(OH)₂  +  2 HCl  ---->  BaCl₂  +  2 H₂O

The measurement and calculation of the amounts of heat exchanged by a system is called calorimetry. The equation that allows calculating these exchanges is:

Q=c*m*ΔT

where Q is the heat exchanged for a body of mass m, constituted by a substance whose specific heat is c, and ΔT is the temperature variation experienced.

In this case:

Given that the solution has the same density as water (1.00 [tex]\frac{g}{mL}[/tex]) then the mass of Ba (OH)₂ and HCl can be calculated as:

[tex]mass Ba(OH)2=70 mL*\frac{1 g}{1 mL}[/tex]

mass Ba(OH)₂=70 g

[tex]mass HCl=70 mL*\frac{1 g}{1 mL}[/tex]

mass HCl=70 g

mass solution = 70 g of Ba(OH)₂ + 70 g of HCl

mass solution = 140 g

Another way to calculate the mass of the solution is:

The total volume is the sum of the individual volumes:

total volume= volume of Ba(OH)₂ + volume of HCl = 70 mL + 70 mL

total volume= 140 mL

Given that the solution has the same density as water (1.00 [tex]\frac{g}{mL}[/tex]) then

[tex]mass solution=140 mL*\frac{1 g}{1 mL}[/tex]

mass solution = 140 g

Then

Q = 140 g* 4.184 [tex]\frac{J}{g*C}[/tex] *4.63° C =2712 J

By reaction stochetry (that is, by the relationships between the molecules or elements that make up the reactants and reaction products) 2 moles of HCl produce 2 moles of H2O.

Then

[tex]moles HCl=70 mL*\frac{1 L}{1000 mL} *\frac{0.700 g}{1 L}[/tex]

moles HCl=0.049

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So

[tex]x=\frac{c*b}{a}[/tex]

In this case:  If 2 moles of H2O are formed if 2 moles of HCl react, how many moles of H2O will be formed if 0.049 moles of HCl react?

[tex]moles H2O=\frac{0.049 moles*2moles}{2 moles}[/tex]

moles H2O=0.049 moles

Now

ΔH/mol H₂O = [tex]\frac{2712 J}{0.049 mol H2O}[/tex]

ΔH/mol H₂O = 55346 J/mol H₂O =55.346 kJ/mol H₂O

Answer: The concentration of [tex]HNO_3[/tex] solution will be 0.094 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]

We are given:

Conversion factor: 1L = 1000 mL

[tex]n_1=1\\M_1=?M\\V_1=0.110L=110mL\\n_2=2\\M_2=0.100M\\V_2=51.9mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 110=2\times 0.100\times 51.9\\\\M_1=0.094M[/tex]

Hence, the concentration of [tex]HNO_3[/tex] solution will be 0.094 M.

The concentration of the HNO3 solution can be found by calculating the moles of Ba(OH)2 used in neutralization, which gives us the moles of HNO3 due to reaction stoichiometry. The HNO3 concentration (molarity) is then found by dividing its moles by its volume.

In the given neutralization reaction, two moles of HNO3 react with one mole of Ba(OH)2. Given that 51.9 mL of 0.100 M Ba(OH)2 is used to completely neutralize the solution, the moles of Ba(OH)2 used can be obtained by using the formula Molarity (M) = Moles/Liter and converting the mL to L. The reaction stoichiometry implies that the moles of HNO3 must be twice that of Ba(OH)2.

Therefore, the total moles of HNO3 can be calculated by doubling the moles of Ba(OH)2. Divide these moles by the volume of the HNO3 solution (0.110 L) to find its molarity.

To summarize, the calculation process involves finding out the moles of Ba(OH)2 used, using this to find the moles of HNO3, and then dividing this by the volume of HNO3 solution to yield the molarity. Please carry out these calculations to get your final answer.

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Answer:

Explanation:

Out of those choices, the chemical that ionizes the most is NaBr. It is like table salt (NaCl) in its properties.  Solutbility 121 grams in 100 mL

You might think "Well if NaBr is such a good conductor, maybe a Metal combined with a non metal is the key and BaCl2 should be next."  And in this case you would be right.  36 grams  will dissolve in 100 mL

C3H7COOH is an acid and it is soluble in water. It is an acid and the last H splits off. It is not quite as good as the top two, but good enough all the same.

Sugar is probably the least soluble but it does form a suspension. It would be at the bottom of the last.

Answer:

The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride

Explanation:

The conductivity of a solution is related to the concentration of ions. The higher the concentration of ions, the higher the conductivity.

Barium chloride is a strong electrolyte, according to the following equation.

BaCl₂(aq) ⇒ Ba²⁺(aq) + 2 Cl⁻(aq)

Each mole of BaCl₂ produces 3 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.60 M in ions.

Sugar (C₆H₁₂O₆) is a non-electrolyte. Thus, the concentration of ions will be zero.

Butanoic acid is a weak electrolyte, according to the following equation.

C₃H₇COOH(aq) ⇄ C₃H₇COO⁻(aq) + H⁺(aq)

Due to its weak nature, the concentration of ions will be lower than 0.20 M.

Sodium bromide is a strong electrolyte, according to the following equation.

NaBr(aq) ⇒ Na⁺(aq) + Br⁻(aq)

Each mole of NaBr produces 2 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.40 M in ions.

The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride

Answer: The half-life of the reaction in 2.24 seconds.

Explanation:

We are given a reaction which follows first order kinetics.

The formula used to calculate the half -life of the reaction for first order kinetics follows:

[tex]t_{1/2}=\frac{0.693}{k}[/tex]

where,

[tex]t_{1/2}[/tex] = half-life of the reaction

k = rate constant of the reaction = [tex]0.310s^{-1}[/tex]

Putting values in above equation, we get:

[tex]t_{1/2}=\frac{0.693}{0.310s^{-1}}\\\\t_{1/2}=2.235sec\approx 2.24sec[/tex]

The rule which is applied for multiplication and division problems is that the least number of significant figures in any number of a problem will determine the number of significant figures in the solution.

In the problem, the least precise significant figures are 3. Thus, the answer will also have 3 significant figures.

Hence, the half-life of the reaction in 2.24 seconds.

Answer: The net ionic equation for the given reaction is [tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)[/tex]

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

[tex]2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)[/tex]

Ionic form of the above equation follows:

[tex]2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^{2-}(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)[/tex]

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow SO_2(g)+H_2O(l)[/tex]

Hence, the net ionic equation for the given reaction is written above.

The net ionic equation for the reaction between hydrochloric acid and potassium sulfite is H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq), following the solubility trends of sulfates and sulfites under standard conditions.

The reaction between excess hydrochloric acid (HCl) and potassium sulfite (K2SO3) is a typical acid-base neutralization reaction. In the initial step, potassium sulfite dissociates into its ions in the aqueous solution:

K2SO3 (aq) → 2K+ (aq) + SO3^2- (aq)

Hydrochloric acid, being a strong acid, also dissociates completely:

HCl (aq) → H+ (aq) + Cl- (aq)

The hydrogen ion from the acid then reacts with the sulfite ion to form sulfuric acid and water, creating a net ionic equation :

2H+ (aq) + SO3^2- (aq) → H2SO3 (aq)

Because of the solubility trends of sulfates and sulfites under standard conditions, the sulfuric acid produced also dissociates into ions:

H2SO3 (aq) → 2H+ (aq) + SO3^2- (aq)

Therefore, the overall net ionic equation is:

H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq)

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Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].

[tex]\rm NH_3[/tex] gains one hydrogen ion to produce the ammonium ion [tex]\rm {NH_4}^{+}[/tex]. In other words, [tex]\rm {NH_4}^{+}[/tex] is the conjugate acid of the weak base [tex]\rm NH_3[/tex].

Both buffer 1 and 2 include

The ammonia [tex]\rm NH_3[/tex] in the solution will react with hydrogen ions as they are added to the solution:

[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].

There are more [tex]\rm NH_3[/tex] in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of [tex]\rm NH_3[/tex] in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.