What is the amount of mass in each given unit of volume?

Chemistry

1 answer:

labwork [276]3 years ago

3 0

Answer:

Each volumn is different and therefore has a specific and unique mass.

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The answer is C. Only metallic alloys are malleable and ductile. The other descriptions can apply to all 3 answers

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ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.

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ICMP Echo Request

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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

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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

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

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

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3 years ago

which statement best describes the equation caco3 2hcl → cacl2 co2 h2o? caco3 is a reactant; it is present before the reaction o

fomenos

Answer: $CaCO_3$ is a reactant; it is present before the reaction occurs.

Explanation:

In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.

$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$