_NH3 + _ 02 →_NO + _ H2O Balanced

The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

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

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

3 years ago

99 96 040Zr + 42He ----> + "42Mo

Ahat [919]

he mass defect of the helium nucleus ⁴He₂ is 0.030377 u

Further explanation

Mass defect means the difference between the mass of particles forming an atom with an atomic mass.

Δm = mass defect ( u )

mp = mass of proton ( u )

me = mass of electron ( u )

mn = mass of neutron ( u )

M = atomic mass ( u )

A = mass number

Z = atomic number

Let us now tackle the problem !

Given :

Unknown :

Δm = ?

Solution :

Learn more

Rutherford’s major achievements : brainly.com/question/1552732

Unit of radius of an atom : brainly.com/question/1968819

Fusion : brainly.com/question/11395223

Answer details

Grade: College

Subject: Physics

Chapter: Nuclear Physics

Keywords: Mass , Defect , Nucleon , Number , Atomic , Proton , Electron , Neutron

4 0

2 years ago

In examples where the volume of a system is under constant pressure, which of

grin007 [14]

Answer:

D

Explanation:

At constant volume, the heat of reaction is equal to the change in the internal energy of the system. ... Most chemical reactions occur at constant pressure, so enthalpy is more often used to measure heats of reaction than internal energy.

6 0

3 years ago

H2(g) + Br2(l) ⇄ 2HBr(g) Kc = 4.8 × 108

elena-14-01-66 [18.8K]

Answer:

1.5 × 10⁻⁹M

Explanation:

1. Equilibrium equation

H₂(g) + Br₂(l) ⇄ 2HBr(g)

2. Equilibrium constant

The liquid substances do not appear in the expression of the equilibrium constant.

$k_c=\dfrac{[HBr(g)]^2}{[H_2]}=4.8\times 10^8M$

3. ICE table.

Write the initial, change, equilibrium table:

Molar concentrations:

H₂(g) + Br₂(l) ⇄ 2HBr(g)

I 0.400 0

C - x +2x

E 0.400 - x 2x

4. Substitute into the expression of the equilibrium constant

$4.8\times 10^8=\dfrac{(2x)^2}{0.400-x}$

5. Solve the quadratic equation

192,000,000 - 480,000,000x = 4x²

x² + 120,000,000x - 48,000,000 = 0

Use the quadratic formula:

$x=\dfrac{-120,000,00\pm\sqrt{(120,000,000)^2-4(1)(-48,000,000}}{2(1)}$

The only valid solution is x = 0.39999999851M

Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M

8 0

3 years ago

Why do experiments need to be repeatable?

maw [93]

If they were not repeatable people would think the experiment is not accurate. If it can be repeated than the data can prove a very valid point.

3 0

3 years ago

Read 2 more answers