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saveliy_v [14]
3 years ago
11

If 10.23 moles of NH3 were produced, how many moles of N2 would be required?

Chemistry
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

the number of moles of N_2 required is 5.12 moles

Explanation:

The computation of the number of moles of N_2 required is shown below:

N₂ +  3H₂ → 2NH₃

Here the nitrogen moles and hydorgen would be compared  

NH₃  :  N₂

2 :  1

10.23  : 1 ÷ 2 × 10.23 = 5.12mol

Hence, the number of moles of N_2 required is 5.12 moles

Therefore the same would be considered and relevant too

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What is the percentage of hydrogen in c2h4
jarptica [38.1K]

Ethylene- C2H4 = 85.7% Carbon and 14.3% Hydrogen


Find the atomic masses for each element and multiply it by the number of atoms in the compound, then add.


C- 12.0 * 2= 24.0


H- 1.00 * 4= 4.00


-----------------------


28.0


Take the masses for each element and divide it by the total mass. Then change the answer to get the percent.


C 24.0 / 28.0= .857 = 85.7%


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8 0
3 years ago
Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?
sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

3 0
3 years ago
Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K.
Elis [28]

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

\Delta G= \Delta H - T * \Delta S

<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>

Calculating the \Delta G= -1267 - 473 K* \Delta S :

\Delta G= -1267 - 473 K* \Delta S

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: \Delta S>0[/tex}Back to this expression: [tex]\Delta G= -1267 - 473 K* \Delta S

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

4 0
3 years ago
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Kipish [7]
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