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Nina [5.8K]
2 years ago
11

AT constant temperature, a gas is compressed from 5.5 L to 1 L. At 1 L, the pressure of the gas is 98.2 kPa. What was the origin

al pressure of this gas? (Use Boyle's Law)
Chemistry
1 answer:
svet-max [94.6K]2 years ago
7 0

We have that  the original pressure of this gas  is mathematically given as

P2=17.854 kPa

<h3>Boyle's law</h3>

Question Parameters:

AT constant temperature, a gas is compressed from 5.5 L to 1 L.

At 1 L, the pressure of the gas is 98.2 kPa

Generally the equation for Boyle's law   is mathematically given as

P1V1=P2V2

Therefore

P1=\frac{P2V2}{V1}\\\\P2=\frac{98.2*1}{5.5}

P2=17.854 kPa

For more information on Pressure visit

brainly.com/question/25688500

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The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

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