A solution is

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Andrew [12]

B. crust, middle, core

8 0

3 years ago

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

A marble rolls off horizontally from the edge of table top 1.50 m above the floor. it strikes the floor 2.0 m from the base of t

GrogVix [38]

a. t=0.553 s

b. vox(horizontal speed) = 3.62 m/s

<h3>Further explanation</h3>

Given

h = 1.5 m

x = 2 m

Required

a. time

b. vo=initial speed

Solution

Free fall motion

a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)

$\tt t=\sqrt{\dfrac{2h}{g} }$

t = √2h/g

t = √2.1.5/9.8

t=0.553 s

b. x=vox.t(horizontal motion)

$\tt x=v_{ox}\times t$

vox=x/t

vox=2/0.553

vox=3.62 m/s

3 0

2 years ago

Which one of these are correct?

Sergio [31]

The correct answer is the one in the middle. Mixing food coloring and water

3 0

3 years ago

How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield

ahrayia [7]

Answer: 100 ml


Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

i) Formula: n = mass in grass / molar mass

ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol


iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃


2) Use the percent yield to calculate the theoretical amount:


65% = 0.65 = actual yield/ theoretical yield =>

theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃

3) Chemical equation:


 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:


3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃

=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume


i) Molarity formula: M = n / V


ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml

3 0

3 years ago

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