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Galina-37 [17]
3 years ago
10

using the water cycle diagram, match the number with the following; condensation runoff [precipitation and evaporation. once you

've identified the process for each number, explain how water moves and changes form in each step (you only need to do this for the 4 basic steps)

Chemistry
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

1. Evaporation

2. Condensation

3. Precipitation

4. Runoff

Explanation:

The water cycle shows how water moves on the earth as it passes through different phases.

Water on land is stored in oceans, rivers and other water bodies. This water coupled with those in plants and animals are turned into vapor through EVAPORATION. Evaporation facilitates the movement of water on the surface into the atmosphere.

In the atmosphere, the vapor condenses on dust particles found up there. The vapors forms a nuclei around the dust particles and water condenses at the saturated vapor pressure. This forms cloud.

As the water collects more and more, gravity forces the water to fall in form of PRECIPITATION. The precipitation can be in form of snow or rainfall.

When precipitation occurs, they move on the surface as SURFACE RUNOFFS. Some of the runoff goes back into oceans and rivers. Others infiltrates into the ground and collects in the ground water pool under the subsurface. Subsurface water can also get into into other water bodies when the water table coincides with the steam level.

The water in these bodies can then go into the cycle again. The sun is the source of energy for this process.

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It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reac
Rashid [163]

It requires 60 mL of 0.20 M sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H_2CO_3), hence the carbonate ions concentration is 0.24M.

Given:

Reaction:

\to \bold{2NaOH + H_2CO_3 \to Na_2CO_3 + 2H_20 }

NaOH volume (V_B) = 60 \ ml  

H_2CO_3 Volume (V_A) = 25\ ml  

NaOH Molarity (C_B) = 0.20\ M

H_2CO_3 moles (n_A) = 1

NaOH moles (n_B) = 2

To find:

H_2CO_3 Molarity (C_A) =?

Solution:

Using the neutralization reaction:  

\to \frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B} \\\\

\to C_B = 0.2\ M \\\\ \to  n_A = 1 \\\\ \to  n_B = 2 \\\\ \to  V_B = 60\ ml \\\\  \to  V_A = 25\ ml

Calculating the C_A:

 \to \frac{C_A \times 25}{0.2  \times 60} =\frac{1}{2} \\\\\to C_A =\frac{1 \times 0.2 \times 60}{2 \times 25}  \\\\\to C_A =\frac{1 \times 2 \times 12}{2 \times 5 \times 10} \\\\ \to C_A =\frac{ 12}{ 5 \times 10} \\\\\to C_A = \frac{6}{5 \times 5} \\\\\to C_A = \frac{6}{25} \\\\\to C_A=0.24\ M

Therefore, the concentration of carbonic acid is "0.24M".

Learn more about the concentration:  

brainly.com/question/9889034

3 0
2 years ago
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ololo11 [35]

Answer:

350 J

Explanation:

The law of conservation of energy states that energy can neither be created nor destroyed but can only be converted from one form to another. This means that in a system, energy is not lost.

In this question, an unlit match contains approximately 1,000 J of chemical energy. When lit, thermal and light energy are emitted i.e. it gets converted to light and heat energy. If the thermal energy emitted was measured to be 400J, and the remaining match still contains 250J of chemical energy, then:

The amount of light energy emitted will be:

Total chemical energy (1000J) - {Remaining chemical energy (250J) + emitted thermal energy (400J)}

= 1000 - (400 + 250)

= 1000 - 650

= 350

Hence, the amount of light energy emitted is 350J

Note that, the amount of energy converted (thermal and light) and remaining chemical energy still equates the total chemical energy in the match.

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Ammonium phosphate ((NH), PO) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (TI,PO.
madreJ [45]

Answer:

22.35 g of  (NH₄)₃PO₄ will be produced by 7.5 g of ammonia.

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17 g NH₃ _________ 1 mol

7.5 g NH₃ __________ x

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3 mol NH₃ __________ 1 mol (NH₄)₃PO₄

0.44 mol NH₃ _________           y

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1 mol (NH₄)₃PO₄ ____________ 149 g

0.15 mol (NH₄)₃PO₄ __________    z

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How many grams of a 22.9% sugar solution contain 68.5 g of sugar?
777dan777 [17]

    The grams  of 22.9 %   sugar  solution that contain 68.5  g  of   sugar  is 299.13 g  of  solution  

   <u><em>calculation</em></u>

 22.9%  means that  there are  22.9 g  of  sugar  in 100 g of solution.

   what about   68.5 g  of sugar

- <em>by cross    multiplication</em>

=[(68.5 g  sugar x 100  g  solution) /22.9  g sugar] =299.13 g  of solution

Nb; <em>g sugar cancel  each other</em>

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3 years ago
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