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Reika [66]
3 years ago
9

13. An aerosol spray can of deodorant with a volume of 0.410 L contains 3.0 g of propane gas (CH3) as propellant. What is the pr

essure in the can at 20°C? VRT 3. og festa​
Chemistry
1 answer:
qaws [65]3 years ago
8 0

Answer: The pressure in the can is 4.0 atm

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = ?

V = Volume of gas = 0.410 L

n = number of moles = \frac{\text {given mass}}{\text {Molar mass}}=\frac{3.0g}{44.1g/mol}=0.068mol

R = gas constant =0.0821Latm/Kmol

T =temperature =20^0C=(20+273)K=293K

P=\frac{nRT}{V}

P=\frac0.068mol\times 0.0820 L atm/K mol\times 293K}{0.410L}=4.0atm

Thus the pressure in the can is 4.0 atm

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((2.0265-2.02)/2.0265) x 100.00
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Answer 0.5

when reporting calculations from sig figs, report the number of sig figs in the least accurate measurement

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((2.0265-2.02)/2.0265) x 100.00

((2.0265/2.0265) - (2.02/2.03)) X 100.00

(1.0000 -0.995) X 100.00

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What would the [OH-] of a solution that has a pOH of 2.7 be?
Brilliant_brown [7]

Answer:

The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.

pOH indicates the concentration of hydroxyl ions [OH-] present in a solution. In this way, pOH is defined as the negative logarithm of the activity of hydroxide ions, that is, the concentration of OH- ions:

pOH= -log [OH⁻]

In this case, pOH has a value of 2.7. Replacing:

2.7= -log [OH⁻]

and solving:

[OH⁻]=10⁻² ⁷

you get:

[OH⁻]≅ 2*10⁻³

<u><em>The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³</em></u>

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