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3 years ago

Name the three types of rain fall

Chemistry

2 answers:

lara31 [8.8K]3 years ago

5 0

Answer:

Hey mate....

Explanation:

This is ur answer....

<h3>The 3 types of rainfall are Convectional, Orographic and Cyclonic.....</h3><h3 /><h3>it is same as Relief Rainfall, Convectional Rainfall and Frontal Rainfall.</h3><h3 />

Hope it helps you!

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Trava [24]3 years ago

4 0

Explanation:

Convectional rainfall.

Orographic or relief rainfall.

Cyclonic or frontal rainfall

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3 years ago

A 53. 5 mL ample of an 5. 4 % (m / v) KBr olution i diluted with water o that the final volume i 205. 0 mL Expre your anwer to t

gogolik [260]

The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

$C_{1}$ $V_{1}$ = $C_{2}$ $V_{2}$ , where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

Keep the values to calculate final concentration.

$C_{2}$ = (53.5 × 5.4)/205.0

Performing multiplication on Right and Side

$C_{2}$ = 288.9/205.0

Performing division on Right Hand Side

$C_{2}$ = 1.4%

Hence, the final concentration is 1.4%.

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brainly.com/question/17206790

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The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

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1 year ago

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Misha Larkins [42]

Answer:

The answer will be C. Variable

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4 years ago

Read 2 more answers

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

5 0

1 year ago

Name the three types of rain fall​

Name the three types of rain fall