What is the acceleration of a car that starts at 0 km/s and speeds up to 50 km/s in 10 seconds?

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

2 years ago

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago

What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?

topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, $R=0.08206\ L-atm/mol-K$

Finding for V,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L$

So, the volume of the gas is 34.55 L.

4 0

3 years ago

Help plzzzzzzzzzzzzz

tatuchka [14]

Answer:

A: element B

B: element A

C: element B

D: element A

Explanation:

decrease in size leads increase in electronegativity because the smaller the size, the closer the shell is to the nucleus. Also, atomic radius decreases to the right and up on the periodic table. Atomic radius increases to the left and down a period. I hope this helps!

6 0

3 years ago

1. How many Chromium atoms are found in 25.8 milligrams of Chromium?

givi [52]

Answer: 51.9961 g/mol, don't know if it helps :)

Explanation:

7 0

3 years ago

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