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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

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

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

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

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

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

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

$A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}$

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

$A=\text{ 50 }\times2^{\frac{-t}{14.3}}$

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

$\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}$

So, after 84 days the P-32 remaining will be 0.85 mg

4 0

1 year ago

Write the formulas of the three singly chlorinated isomers formed when 2,2-dimethylbutane reacts with Cl2 in the presence of lig

Tems11 [23]

The molecule with same molecular formula but different arrangement of atoms is said to be an isomer.

When 2,2-dimethylbutane reacts with chlorine in the presence of light gives three isomers that is $CH_{3}C(CH_{3})_{2}CHClCH_{3}$ (3-chloro-2,2-dimethylbutane), $ClCH_{2}C(CH_{3})_{2}CH_{2}CH_{3}$ (1-chloro-2,2-dimethylbutane) and $ClCH_{2}CH_{2}C(CH_{3})_{2}CH_{3}$ (1-chloro-3,3-dimethylbutane).

In above case, the molecular formula of all isomers are same i.e. $C_{6}H_{13}Cl$ but chlorine is arranged in different positions of carbon. Thus, results isomers.

The reaction is shown in the image.

3 0

3 years ago

What type of energy results from the

jeka94

Answer:

C fghhtrsfeagutyrwraqedf

3 0

3 years ago

An airplane starts at rest and Excelerator down the runway for 20 seconds. At the end of the runway, its velocity is 80 m/s Nort

masha68 [24]

According to the formula you have given us to work with . . .

1). The airplane's acceleration is

(80 m/s north - zero) / (20 sec) = 4 m/sec^2 north

2). For the cyclist:

(V-final - zero) / 20sec = 0.5 m/s^2 south

Multiply each side by 20s : V-final = 0.5 m/s^2 south x (20sec) =

10 m/s south

8 0

3 years ago