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gtnhenbr [62]
3 years ago
12

Select True or False: The equilibrium constant for the chemical equation 2NO(g) O2(g) 2NO2(g) is two times the equilibrium const

ant for the chemical equation NO(g) 1/2O2(g) NO2(g).
Chemistry
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:

False

Explanation:

The first reaction is;

NO(g) + 1/2O2(g) ---->NO2(g)

K= [NO2]/[NO] [ O2]^1/2

The second reaction is;

2NO(g) + O2(g) ---->2NO2(g)

K'= [NO2]^2/[NO]^2 [O2]

It now follows that;

K'= K^2

Hence the statement in the question is false

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A sample of water at 100°C is converted to steam after absorbing 820 kJ of heat. How grams of H2O are contained in the sample? A
Varvara68 [4.7K]
Since water is already at 100<span>°C all the energy is used to evaporate it. 
Now we can calculate how many </span>mols of water are evaporated with 820kJ.
N= \frac{820}{41} =20 mol
We calculated that we got 20 mols of water evaporated. Now, all we have to do is find how many grams is a mol of water. Molar mass of water is <span>20.16 g/mol.
</span>The final answer is:
m=20*20.16=403.2g




7 0
3 years ago
What happens when acetylene reacts with silver powder?​
horrorfan [7]
It dissolves I think I know I am expert and. Ute
6 0
2 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction
const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>

KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>

8 0
2 years ago
A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical
LuckyWell [14K]

The empirical formula :

C₁₀H₁₆N₄SO₇

<h3>Further explanation</h3>

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

= 2/18 x 2.75 g

= 0.305

mass N = 1.06

mass S :

= 32/64 x 1.23

= 0.615

mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g

Mol ratio :

= C : H : N : S : O

= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16

= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇

3 0
3 years ago
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