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3 years ago

what is the new pressure acting on a 2.5 l balloon if its original volume was 5.8 l at 3.7 ATM of pressure

Chemistry

1 answer:

Ivan3 years ago

8 0

Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 3.7 atm

$P_2$ = final pressure = ?

$V_1$ = initial volume = 5.8 L

$V_2$ = final volume = 2.5 L

Now put all the given values in the above equation, we get:

$3.7atm\times 5.8L=P_2\times 2.5L$

$P_2=8.6atm$

Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.

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How many atoms of San are in 0.796 moles of this element

xenn [34]

Answer:

4.8× 10²³ atoms

Explanation:

Given data:

Number of moles of San element = 0.796 mol

Number of atoms present = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.796 mol × 6.022 × 10²³ atoms / 1mol

4.8× 10²³ atoms

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2 years ago

This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press

Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀ = - kt

thus:

(pPH₃ )t = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

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