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3 years ago

PLEASE HELP ME QUICK

Chemistry

1 answer:

Mama L [17]3 years ago

6 0

Answer:

b. decreases

C. neither a physical change nor a chemical change

light work

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Polar air masses bring warm temperature and have lower air temperature? T or F

snow_tiger [21]

Answer:

t? im pretty sure have a good day

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3 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

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

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

4 0

3 years ago

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

77.In the early days of automobiles, illumination at night was provided by burning acetylene, C2H2. Though no longer used as aut

natta225 [31]

Answer:

554.86kj

Explanation: Since 1 mole of CaC2=15.14kj yield 1mole of C2H2

The enthalpy change of H2O is 2*285=570

570+-15.14=554.86kj

Hence Hp is 554.86kj

He=Hp

3 0

3 years ago

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