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2 years ago

PLEASE ASNWR ASAP

Chemistry

1 answer:

Vikki [24]2 years ago

4 0

34.95 atm

lol i hope i’m not too late

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What is the approximate temperature of 1.4 moles of a gas with a pressure of 3.25 atmospheres in a 4.738-liter container

attashe74 [19]

Answer:

134K

Explanation:

Using the ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas constant (0.0821 Latm/Kmol)

T = temperature (K)

Based on the information provided, n = 1.4moles, P = 3.25atm, V = 4.738L, T = ?

3.25 × 4.738 = 1.4 × 0.0821 × T

15.3985 = 0.11494T

T = 15.3985/0.11494

T = 133.969

Approximately;

T = 134K

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2 years ago

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t

Serga [27]

Explanation:

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

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Answer:

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