Is au an atomic element, molecular element, molecular compound or an ionic compound?

To produce energy, nuclear power plants use a process called...

vovangra [49]

I think it’s b but I could be wrong

5 0

3 years ago

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

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

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

3 years ago

Read 2 more answers

Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o

stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of butane

$\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles$

b) moles of oxygen

$\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles$

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

According to stoichiometry :

2 moles of butane require 13 moles of $O_2$

Thus 0.09 moles of butane will require = $\frac{13}{2}\times 0.09=0.585moles$ of $O_2$

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0

3 years ago

What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

nirvana33 [79]

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

8 0

3 years ago

Why does a hot air balloon rise when the air inside it is heated

Nastasia [14]

hot air rises up in the balloon and lifts the balloon up

7 0

3 years ago