Which statement is true about this reaction? mc016-1.jpg

Chemistry

1 answer:

Julli [10]3 years ago

8 0

C because, its products is heavier than each of its reactants

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The following chemical reaction occurs in a basic solution.

Dafna1 [17]

Answer:

6 moles of electrons

Explanation:

Let us consider the species NO3− and ClO−. The NO3− is oxidized to NO the oxidation number of nitrogen is decreased from +5 to +2.

The oxidation number of chlorine is increased from +1 to +3. This implies that six electrons were transferred in the balanced reaction equation shown in the question. Hence the answer.

3 0

2 years ago

what is the molarity of a RbOH solution if 60.0 mL of the solution is neutralized by 52.8 mL of a 0.5M HCl solution (Hint: Ma x

Dahasolnce [82]

RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL
Mb = 0.44 M
molarity of base - 0.44 M

7 0

2 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$