A] 1 N2 + 3 H2 → 2 NH3

Chemistry

1 answer:

Vilka [71]3 years ago

8 0

I haven't done these in some time, so I'm not sure if they are 100% right.

1) 6.3 moles of H2( 2 mol of NH3 / 3 mol of H2)= 4.2 mol of NH3

6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3

2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3

2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3

3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2

4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2

10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2

30.3 g = .0303 liters of H2

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Plz help, I will give brainliest.

baherus [9]

Answer: The value of $K_c$ is 2

Explanation:

Moles of $N_2O_4$ = 1.0 mole

Volume of solution = 1.00 L

Initial concentration of $N_2O_4$ = $\frac{1.0mol}{1.00L}=1.0M$

Equilibrium concentration of $NO_2$ = $\frac{1.0mol}{1.00L}=1.0M$

The given balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial conc. 1.0 M 0 M

At eqm. conc. (1.0-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Given : 2x = 1.0

x= 0.5

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

$K_c=\frac{(1.0)^2}{(1.0-0.5)}$

$K_c=2$

Thus the value of $K_c$ is 2

4 0

3 years ago

Draw every stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane. Use wedge-and-dash bonds for the substituent groups, and be s

Blizzard [7]

Answer:

See figure 1

Explanation:

In this question, we have to find first the chiral carbons on 1,2-difluoro-1,2-dimethylcyclopentane. Carbons 1 and 2. Then, using the Cahn–Ingold–Prelog priority rules. We can draw all the stereoisomers. If we have 2 chiral carbons we will have ( $2^n=2^2=4$ ) 4 isomers. With this in mind we can have the R,R isomer; the S,S isomer; the S,R isomer, and the R,S isomer. We have to remember that: