All categories

2 years ago

Which of the following laboratory procedures best illustrates the law of conservation of mass?

Chemistry

2 answers:

77julia77 [94]2 years ago

4 0

Answer is: Synthesizing 36 g of H2O plus leftover reactants by combining 4 g of H2 and 32 g of O2.

m(H₂) + m(O₂) = m(H₂O).

4 g + 32 g = 36 g.

Conservation of mass: during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.

Balanced chemical reaction: 2H₂ + O₂ → 2H₂O.

m(H₂) = 4 g; mass of hydrogen.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

m(O₂) = 32 g.

n(O₂) = 32 g ÷ 32 g/mol.

n(O₂) = 1 mol; amount of oxygen.

n(H₂) : n(O₂) = 2 mol : 1 mol according to balanced chemical reaction.

zavuch27 [327]2 years ago

4 0

Synthesizing 36 g of H2O because the mass of the reactants add up to the mass of the product

You might be interested in

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

2 years ago

What is the ph of a 0.014 M HCL solution?

Allushta [10]

Answer:

pH $= 1.853$

Explanation:

For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.

$[H_3O^+] = [HCl] = 0.014$

pH $= -log [H^+] = -log [H_3O^+]$

Substituting the available values in above equation, we can say that the pH of the solution is equal to

$- log (0.014)$

pH $= 1.853$

pH of a $0.014$ M HCL solution $= 1.853$

8 0

2 years ago

In a colloid, solution, or suspension, particles are dispersed throughout the mixture. What is the order of these three types of

a_sh-v [17]

1) solution
2) colloid
3) suspension

4 0

3 years ago

Read 2 more answers

Is sodium oxide an acidic oxide or a basic oxide

valkas [14]

Answer:It a Basic oxide

Plz trust Me

Explanation:

5 0

2 years ago

Read 2 more answers

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago