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sweet-ann [11.9K]
3 years ago
10

Which step will decrease the pressure of a gas inside a closed cubical container? increasing the number of moles of gas decreasi

ng the volume of the container increasing the speed of the gas particles decreasing the temperature inside the container
Chemistry
2 answers:
Jobisdone [24]3 years ago
8 0

Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

Explanation:

According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.

(Pressure)\propto (Temperature)

At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.

Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

Brums [2.3K]3 years ago
7 0

decreasing the temperature inside the container

hope this helps!

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Nitrogen (N) is much more electronegative than hydrogen (H). Which of the following statements is correct about the atoms in amm
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Explanation:

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Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an
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Answer: the correct option is 2, 3, 4, 1.

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4)Lucinda Topper

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While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

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What is the equation for figuring out the Atomic Mass
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Atomic mass = mass of protons + mass of neutrons

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In an element atomic mass of an atom can be calculated by adding the mass of protons and neutrons

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The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
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3,964 years.

Explanation:

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  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

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<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

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[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

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