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3 years ago

How does most of the water in the water cycle move from lakes and rivers directly back into the atmosphere?

Chemistry

2 answers:

olga2289 [7]3 years ago

8 0

It happens through precipitation when water evaporates from the suns rays and rises into the atmosphere

AnnyKZ [126]3 years ago

7 0

It goes from lakes and as the temp rises, it evaporates into clouds. from there, the clouds get heavy, and it produces rain, and the rain forms rivers and then back into lakes.

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A hydrate of zinc nitrate has the formula Zn(NO3)2 . xH2O. If the mass of 1 mol of anhydrous zinc nitrate is 63.67% of the mass

weqwewe [10]

MM Zn(NO₃)₂ = 189.36 g/mol

mass 1 mol Zn(NO₃)₂ = 189.36 g

mass hydrate = 100 / 63.67 x 189.36 = 297.409 g

mass 1 mol hydrate = 297.409 g

MM hydrate = 297.409 g/mol

MM hydrate = MM Zn(NO₃)₂ + MM xH₂O

297.409 = 189.36 + x(18)

x = 6

6 0

2 years ago

What is mole in chemistry?

Zigmanuir [339]

Answer:

the animal

Explanation:

a rodent thas in the ground

6 0

3 years ago

A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a

KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

Partial pressure of methane (pCH₄): 431 mmHg

Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0

3 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

3 years ago

The mass of the electron is..

musickatia [10]

The answer to the question is b

5 0

3 years ago

Read 2 more answers