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Bess [88]
3 years ago
7

The water cycle is the movement of water from Earth's surface to the atmosphere and back to Earth's surface. Why do many living

organisms depend on the water cycle?
A.
The water cycle evenly distributes water across Earth's surface.
B.
The water cycle renews the supply of fresh water on land.
C.
The water cycle removes salts and pollutants from the ocean.
D.
The water cycle constantly adds water to Earth's total supply.
Chemistry
2 answers:
katrin2010 [14]3 years ago
8 0
The water cycle renews the supply
sladkih [1.3K]3 years ago
7 0
The water cycle renews water supplies by moving water from the oceans, as freshwater, to land
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I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the
stiv31 [10]
Molarity of solution is mathematically expressed as,
M = \frac{x\text{weight of solute(g)}}{\text{Molecular weight X Volume of solution(l)}}

We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l

Also, we know that molecular weight of sucrose = 342.3 g/mol
 ∴M = \frac{x\text{211.4}}{\text{342.3 X 0.08849}}
       = 6.979 M

Thus, molarity of solution is 6.979 M
8 0
3 years ago
What are the Number of molecules of 14 grams of NO2
algol [13]

6.022 x 1023 atoms are in 14 grams of NO2

5 0
3 years ago
What is the most common isotope of lithium?
Advocard [28]
Lithium-7 is the most common isotope of lithium.
7 0
3 years ago
Name the separation technique which could be used to separate mud and water without heating​
Westkost [7]

Answer:

Decantation

Explanation:

Decantation is a process to separate

mixtures by removing a liquid layer that

is free of a precipitate, or the solids

deposited from a solution. The purpose

may be to obtain a decant (liquid free

from particulates) or to recover the

precipitate.

Decantation relies on gravity to pull

precipitate out of the solution, so there is

always some loss of product, either from

the precipitate not fully falling out of the

solution or from liquid remaining when

separating it from the solid portion.

The Decanter

A piece of glassware called a decanter is

used to perform decantation. There are

several decanter designs. A simple

version is a wine decanter, which has a

wide body and a narrow neck. When

wine is poured, solids stay in the base of

the decanter.

In the case of wine, the solid is usually

potassium bitartrate crystals. For

chemistry separations, a decanter may

have a stopcock or valve to drain the

precipitate or dense liquid, or it may

have a partition to separate fractions.

Use of Alum and filtration process can also be administered for further purification

8 0
3 years ago
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