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3 years ago

What type of air masses might make up the cold air mass in a cold front

Chemistry

2 answers:

Elena L [17]3 years ago

4 0

Answer:

continental polar air masses

Explanation:

forms over Canada, move down from North Pole, bring bitterly cold and dry air to the northern United States during the winter.

salantis [7]3 years ago

3 0

The air mass in a cold front is air masses that make up cold

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Sloan [31]

Answer:

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Explanation:

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2 years ago

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alekssr [168]

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2 years ago

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Answer:

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2 years ago

Calculate the molar concentration of the polycyclic aromatic hydrocarbon (pah), anthracene (178.23 g/mol), that was found in a w

tester [92]

solution:

Calculate the molar concentration of the polycyclic aromatic hydrocarbon(PHA)(178.23g/mol),that was found in a well water sample at a concentration of 6.21ppb.Assume the density of the water is 1.00mg/mL $ppb corresponds to 1\mu g in 1L of sample \\the 6.21 ppb is 6.21\mu g in 1L of sample\\6.21\mu g PHA\times\frac{1mg PHA}{1000\mu g PHA}\times\frac{1g PHA}{1000\mu mg PHA}\times\frac{1mol. PHA}{178.23g PHA}\\3.48\times10^-8mol PHA\\and if we have 1L of solution\\m=\frac{moles}{v(L)}=\frac{3.48\times10^-8mol}{1L}\\3.348\times^10-8 M$

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2 years ago

Read 2 more answers

Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilib

DaniilM [7]

Answer:

[C] = 0.4248M

Explanation:

A + B ⇄ 2C

C(i) 1.68M 1.68M 0.00

ΔC -x -x +2x

C(eq) 1.68-x 1.68-x 2x

Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²

Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²

=> 2x/1.68 - x = 0.2895

=> 2x = 0.2895(1.68 - x)

=> 2x = 0.4863 - 0.2895x

=> 2x + 0.2895x = 0.4863

=> 2.2895x = 0.4863

=> x = 0.4863/2.2895 = 0.2124

[C] = 2x = 2(0.2124)M = 0.4248M in 'C'

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2 years ago

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