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2 years ago

Which quantity is held constant when working with Boyle’s, Charles’s, and Gay-Lussac’s laws?

Chemistry

2 answers:

choli [55]2 years ago

8 0

The quantity that is held constant when working with Boyle's, Charles's, and Gay-Lussac's laws are moles.

i Just passed the unit test.

QveST [7]2 years ago

6 0

<h2>Answer:</h2>

The correct answer is the amount of gas or number of moles of gas.

<h3>Explanation:</h3>

Boyle s law:

Boyle's law describes that at constant temperature and amount of gas, the pressure and volume of gas are inversely proportional to each other.

Charles's law:

It described that at constant pressure and number of moles of gas, the temperature and volume of gas are directly proportional to each other.

Gay-Lussac's law:

According to this law at constant volume and amount of gas, the temperature and pressure are directly proportional.

Hence according to above described 3 laws, the moles or mount of gas remains constant.

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Can you pls tell me the net ironic equation of H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)

almond37 [142]

Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

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2 years ago

what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C

svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

$\frac{V}{T}=k$

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 0.730 L
T1= 28 °C= 301 °K (0°C= 273°K)
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.730 L}{301K}=\frac{V2}{273K}$

Solving:

$V2=273K*\frac{0.730L}{301K}$

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

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