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Alekssandra [29.7K]
3 years ago
5

A gas has a pressure of 0.12 atm at 294 K. What is the new pressure at 273 K?​

Chemistry
1 answer:
seropon [69]3 years ago
5 0

Answer:

P₂ = 0.11 atm

Explanation:

Given data:

Initial pressure = 0.12 atm

Initial temperature = 294 K

Final pressure = ?

Final temperature = 273 K

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

0.12 atm / 294 K = P₂/273 K

P₂ = 0.12 atm × 273 K / 294 K

P₂ = 32.76 atm. K /294 K

P₂ = 0.11 atm

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Answer is 2KClO3 3O2 + 2KCl
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If energy was added to solid, what state would it change to
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Explanation:

The answer is C Liquid

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For each row in the table below, decide whether the pair of elements will form a molecular or ionic compound. If they will, then
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Answer:

\begin{array}{cccll}\textbf{Element 1} & \textbf{ Element 2} &\textbf{Compound?} &\textbf{Formula} &\textbf{Type}\\\text{Ar}&\text{Xe} &\text{No} &\text{None}&\text{Neither}\\\text{F}& \text{Cs} &\text{Yes} &\text{CsF} &\text{Ionic}\\\text{N} &\text{Br} &\text{Yes} & \text{NBr}_{3}&\text{molecular} \\\end{array}

Explanation:

You look at the type of atom and their electronegativity difference.

If ΔEN <1.6, covalent; if ΔEN >1.6, ionic

Ar/Xe: Noble gases; no reaction

F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic

N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.

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3 years ago
Is a glass window good reflector of heat??
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Answer:

Yes, Is a glass window good reflector of heat.

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A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man
kap26 [50]
<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

  • Mass of NaBr sample is 11.97 g
  • % composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

  • Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

                         = 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0
3 years ago
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