All categories

2 years ago

A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm. (Boyle)

Chemistry

1 answer:

vlabodo [156]2 years ago

6 0

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{4.31 \times 0.755}{1.25} = \frac{3.25405}{1.25} \\ = 2.60324$

We have the final answer as

Hope this helps you

You might be interested in

If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?

leva [86]

Answer:

14.434 r.a.m.

Explanation:

The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

5 0

2 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

What's the formula of mercury (I) dioxonitrate III

steposvetlana [31]

<h2>Answer:</h2>

<h2>Explanation:</h2>

<h3>Formula of mercury ( 1 ) dioxonitrate 111 is Hg2 ( NO2 )2//</h3>

7 0

2 years ago

Calculate the whole-number ratio of staggered to eclipsed conformers that are present at room temperature. Use the Actual G˚ bar

natka813 [3]

Answer:

The ratio of staggered to eclipsed conformers is 134

Explanation:

It is possible to determine the ratio of staggered to eclipsed conformers of a reactant, using the equilibrium:

Staggered ⇄ Eclipsed

Keq = [Eclipsed] / [Staggered]

That means Keq is equal to the ratio we need to find:

Using:

G˚= -RTln Keq

<em>Where G° = -12133.6J/mol</em>

<em>R is gas constant: 8.314J/molK</em>

<em>T is absolute temperature: 298K</em>

<em />

-12133.6J/mol= -8.314J/molK*298K ln Keq

4.8974 = ln Keq

134 = Keq = [Eclipsed] / [Staggered]

<h3>The ratio of staggered to eclipsed conformers is 134</h3>

3 0

2 years ago

What does expand mean in science

borishaifa [10]

Spread out, usually in every direction.

5 0

2 years ago

Read 2 more answers