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alexgriva [62]
3 years ago
15

Which of the following accurate describes the atmospheric conditions required for the formation of clouds

Chemistry
1 answer:
anygoal [31]3 years ago
4 0

Answer:

To make a cloud, air needs only to be cooled to saturation and beyond. ... Usually the water vapor amounts in air are less that this amount. When this is the case, we describe the air as being unsaturated. If we add more vapor to such air, we could reach the maximum for the temperature

Explanation:

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Cylinder of air at 1.5 atm of pressure is kept at room temperature while a piston compresses the air from 40 l down to 10 ml. wh
djyliett [7]

The new pressure, P₂ is 6000 atm.

<h3>Calculation:</h3>

Given,

P₁ = 1.5 atm

V₁ = 40 L = 40,000 mL

V₂ = 10 mL

To calculate,

P₂ =?

Boyle's law is applied here.

According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.

                                        PV = constant

Therefore,

P₁V₁ = P₂V₂

Put the above values in the equation,

1.5 × 40,000 = P₂ × 10

P₂  = 1.5 × 4000

P₂  = 6000 atm

Therefore, the new pressure, P₂ is 6000 atm.

Learn more about Boyle's law here:

brainly.com/question/23715689

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8 0
2 years ago
Which cohesive forces are the strongest in water?
jeyben [28]

When cohesive force is stronger than the adhesive force: concave up meniscus, water forms droplets on surface

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3 years ago
How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?
givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

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Explanation:

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