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Is nitrogen gas a pure substance or a mixture

Verizon [17]

Its a mixture because nitrogen gas is air. Air is homogeneous mixture

3 0

3 years ago

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40

Ronch [10]

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Explanation:

For potassium permanganate:

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

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

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol$

For hydrochloric acid:

To calculate the moles of hydrochloric acid, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol$

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

$2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O$

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = $\frac{2}{16}\times 0.27=0.033moles$ of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = $\frac{5}{16}\times 0.27=0.0843moles$ of chlorine gas.

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

$PV=nRT$

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $40^oC=[40+273]K=313K$

Putting values in above equation, we get:

$1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L$

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

3 0

4 years ago

How can you increase the gravitational potential energy (GPE) of an empty shoe box on the bottom shelf of a bookcase?

Lelu [443]

Answer:someone answer 4 b

Explanation:

123

7 0

3 years ago

Calculate the freezing point of a solution that contains 8.0 g of sucrose (c12h22o11) in 100 g of h2o. kf for h2o = 1.86c/m

Sever21 [200]

Answer is: the freezing point of the solution of sucrose is -0.435°C.
m(H₂O) = 100 g ÷ 1000 g/kg = 0.1 kg.
m(C₁₂H₂₂O₁₁) = 8.0 g.
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 8.0 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0233 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0233 mol ÷ 0.1 kg.
b(solution) = 0.233 m.
ΔT = b(solution) · Kf(H₂O).
ΔT = 0.233 m · 1.86°C/m.
ΔT = 0.435°C.
Tb = 0°C - 0.435°C = -0.435°C.

3 0

3 years ago

Calculate the molality of a 20.0 percent by weight aqueous solution of nh4cl. (molecular weight: nh4cl = 53.5)

cestrela7 [59]

Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.

8 0

4 years ago

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