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3 years ago

Explain why heating a contained gas that is held at a constant volume increases its pressure.

Chemistry

2 answers:

user100 [1]3 years ago

8 0

Answer:

its not able to release the pressure

Explanation:

its not able to release pressure

Evgesh-ka [11]3 years ago

3 0

Gay Lussac's Law - states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure.

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Food is the only limiting factor that keeps populations from growing too large true or false?

topjm [15]

FALSE

There are other limiting factors like lack of space, diseases, and com petition.

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3 years ago

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What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised

musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

Initial pressure (P₁): 0.985 atm

Initial volume (V₁): 3.65 L

Final pressure (P₂): 861.0 mmHg

Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

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2 years ago

How many photons are produced in a laser pulse of 0.364 J at 477 nm?

Luda [366]

Answer:

1.00 × 10¹⁸

Explanation:

1. Calculate the energy of one photon

The formula for the energy of a photon is

E = hc/λ

h = 6.626 × 10⁻³⁴ J·s; c = 2.998 × 10⁸ m·s⁻¹

λ = 477 nm = 477 × 10⁻⁹ m Insert the values

E = (6.626 × 10⁻³⁴ × 2.998× 10⁸)/(477 × 10⁻⁹)

E = 4.165× 10⁻¹⁹ J

2. Calculate the number of photons

Divide the total energy by the energy of one photon.

No. of photons = 0.418 × 1/4.165 × 10⁻¹⁹

No. of photons = 1.00 × 10¹⁸

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3 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

Answer: 4.999 moles of excess reactant will be left over.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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2 years ago

What tendency is the standard reduction potential of an element based on?

Y_Kistochka [10]

I think the answer is B :)

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3 years ago

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