What does the Law of Conservation of Energy say?

Small quantites of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.

ira [324]

<u>Answer:</u> The mass or zinc reacted is 0.624 grams.

<u>Explanation:</u>

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm (Conversion factor: 1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

$p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm$

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

$PV=nRT$

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = $30^oC=[30+273]K=303K$

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

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol$

The chemical equation for the reaction of zinc and hydrochloric acid follows:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, $9.55\times 10^{-3}mol$ of hydrogen gas is produced from = $\frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol$ of zinc metal

To calculate the mass of zinc metal, we use the equation:

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

Molar mass of zinc = 65.38 g/mol

Moles of zinc = $9.55\times 10^{-3}$ moles

Putting values in above equation, we get:

$9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g$

Hence, the mass or zinc reacted is 0.624 grams.

4 0

3 years ago

given the mass lost during the reaction when the alka seltzer is the limiting reagent and that the molar mass of carbon dioxide

Vadim26 [7]

Alka-seltzer is an effervescent antacid which neutralizes the acid of the stomach by generating CO2. The chemical formula of alka-seltzer is C16H17NaO14. In this case, if we are given the mass of the antacid used, then by stoichiometry, mass alka-seltzer * (1/ molar mass of alk-selz) * (1 mol Co2 / 1 mol alk-seltz) * (44 g/molCo2). That is how you get the mass of Co2 produced.

5 0

3 years ago

7. How many liters of NH3, at STP will react with 10.6 g O2 to form NO2 and water? 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) *

Alexus [3.1K]

7) Answer is: c. 4.24 L.

Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

m(O₂) = 10.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 10.6 g ÷ 32 g/mol.

n(O₂) = 0.33 mol; amount of oxygen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

V(O₂) = n(O₂) · Vm.

V(O₂) = 0.33 mol · 22.4 L/mol.

V(O₂) = 7.42 L; volume of oxygen.

From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.

n(NH₃) = 4 · 0.33 mol ÷ 7.

n(NH₃) = 0.188 mol; amount of ammonia.

V(NH₃) = 0.188 mol · 22.4 L/mol.

V(NH₃) = 4.24 L.

8) Answer is: a. 3.7 g.

Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).

m(P₄) = 3.4 g; mass of phosphorous.

n(P₄) = m(P₄) ÷ M(P₄).

n(P₄) = 3.4 g ÷ 123.9 g/mol.

n(P₄) = 0.0274 mol; limiting reactant.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.

n(PH₃) = 4 · 0.0274 mol.

n(PH₃) = 0.1096 mol.

m(PH₃) = 0.1096 mol · 34 g/mol.

m(PH₃) = 3.726 g.

9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.

Percent yield = actual yield / theoretical yield.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

For example:

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

3 0

3 years ago

Describe two ways you think a sound wave could be stopped or could change change direction.

Alla [95]

Answer:

Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the obstacle, and transmission (accompanied by refraction) into the obstacle or new medium

8 0

3 years ago

Read 2 more answers

A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming

ahrayia [7]

Answer:

sorry but i cant

Explanation:

8 0

3 years ago