All categories

3 years ago

At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen:_________.

Chemistry

1 answer:

fgiga [73]3 years ago

7 0

Answer:

e. 4.1 x 10^−4

Explanation:

For the reaction;

2N2O5(g) → 4NO2(g) + O2(g)

The rate of formation is given as;

(1 / 4) Δ [O2] / Δt = (1 / 2 )Δ [N2O5] / Δt

Δ [O2] / Δt = 8.1 x 10^−4 M/s

Inserting into he equation, we have;

(1/4) (8.1 x 10^−4 ) = (1/2) (Δ [N2O5] / Δt)

2.025 x 10^−4 = (1/2) (Δ [N2O5] / Δt)

Δ [N2O5] / Δt = 2 * 2.025 x 10^−4

Δ [N2O5] / Δt = 4.1 x 10^−4

Correct option is option E.

You might be interested in

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

What is changed when matter undergoes a physical change?

Furkat [3]

I think its the mass that changes but im not sure

3 0

3 years ago

You prepare a 2nd solution (solution B) by pipetting 10.00 ml of your 1st solution (solution A) into a 50.00 ml volumetric flask

elena-14-01-66 [18.8K]

Answer:

Concentration solution A was 0.5225 M

Explanation:

10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B

According to laws of dilution- $C_{A}V_{A}=C_{B}V_{B}$

where, $C_{A}$ and $C_{B}$ are concentration of solution A and B respectively

$V_{A}$ and $V_{B}$ are volumes of solution A and B respectively

Here $C_{B}$ = 0.1045 M, $V_{B}$ = 50.00 mL and $V_{A}$ = 10.00 mL

Hence, $C_{A}=\frac{C_{B}V_{B}}{V_{A}}=\frac{(0.1045M\times 50.00mL)}{10.00mL}=0.5225M$

So, concentration solution A was 0.5225 M

8 0

3 years ago

Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia

Wittaler [7]

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

N₂ : NH₃

1 : 2

1.375 : 2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia are produced from 38.5 g of nitrogen.

4 0

3 years ago

What changes are observed on heat capacity ratio and output temperatures by increasing the specific heat capacity of both the fl

BlackZzzverrR [31]

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

8 0

3 years ago