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3 years ago

given the balanced equation representing a reaction 2NO+O2->2NO2+energy the mole ratio of NO to NO2 IS

1 answer:

4 0

The problem tells us that the equation is already balanced, so we know that we don't have to change any of the coefficients in front of the compounds. Therefore, we can find the ratio by looking at the coefficients of NO and NO²:

2NO : 2NO²

This can be simplified to 1 : 1. Therefore, it is a 1 to 1 ratio.

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When the volume of a gas is changed from 11.5 cm3 to ?cm the temperature will change from 415 K to 200 K

Sonja [21]

Assuming that the gas acts like an ideal gas, we can calculate for the final volume using the ideal gas law:

PV = nRT

Where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature

Assuming that P, n, and R are constant throughout the process, we can define another constant K:

V / T = K where K = nR / P

Equating the initial and final states:

Vi / Ti = Vf / Tf

Substituting the given values:

11.5 cm^3 / 415 K = Vf / 200 K

Vf = 5.54 cm^3

4 0

3 years ago

7.32 moles of hydrogen reacts with 48.97 grams of nitrogen, how many moles of ammonia is produced?

faust18 [17]

3.5 moles of ammonia (NH₃) are produced

Explanation:

We have the following chemical reaction where hydrogen (H₂) reacts with nitrogen (N₂) to produce ammonia (NH₃):

3 H₂ + N₂ → 2 NH₃

number of moles = mass / molecular weight

number of moles of N₂ = 48.97 / 28 = 1.75 moles

We see from the chemical reaction that 1 mole of N₂ will react with 3 moles of H₂, so 1.75 moles of nitrogen will react with 3 × 1.75 = 5.25 moles of H₂. We have 7.32 moles of H₂, a quantity more of what is needed, so the limiting reactant is N₂.

Knowing this we devise the following reasoning:

if 1 mole of N₂ produces 2 moles of NH₃

then 1.75 moles of N₂ produces X moles of NH₃

X = (1.75 × 2) / 1 = 3.5 moles of NH₃

Learn more about:

limiting reactant

brainly.com/question/7144022

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7 0

3 years ago

How many moles of air must escape from a 10-m  8.0-m  5.0-m room when the temperature is raised from 0c to 20c? assume the p

slega [8]

Data:

p = 1 atm
V = 10 m * 8 m * 5 m = 400 m^3 = 400,000 liter

To = 0 + 273.15K = 273.15K
Tf = 20 + 273.15K = 293.15K

No - Nf =?

2) Formula

pV = NRT => N = pV / (RT)

3) solution

No = pV / (RTo)

Nf = pV / (RTf)

=> No - Nf = [pv / R] [ 1 / To - 1 / Tf ]

=> No - Nf = [1atm*400,000liter / 0.0821 atm*liter/K*mol ] [ 1 / 273.15 - 1 / 293.15]

No - Nf = 1216.9 moles ≈ 1217 moles

Answer: 1217 moles

3 0

2 years ago

In the following question, select the odd word from the given alternatives.

mote1985 [20]

Answer:

Explanation:

A B and D are all fosil fuses c is not

5 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

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

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

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

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago