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Svetllana [295]
3 years ago
6

pleeeaaaseeeee answer asappppppppp!!!!!!!!!!! Explain the processes of evaporation and condensation using the image above

Chemistry
2 answers:
goldfiish [28.3K]3 years ago
5 0
In the lake that the rivers lead, water molecules evaporate into the sky and form clouds. In the sky, these water droplets condense and form clouds that will eventually rain.
stepan [7]3 years ago
3 0
The water evaporates and then, the clouds form condensation, because the gas becomes water, and falls as rain. 
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Calculate the extinction coefficient where the concentration is in mg/ml and the path length is 1 cm. What dilutions of the stoc
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Complete Question

The complete question is show on the first uploaded image

Answer:

This is shown on the second,third , fourth and fifth image

Explanation:

This is shown on the second,third , fourth and fifth image

4 0
3 years ago
. Which of these is an example of a scientific observation? *
Yakvenalex [24]

Answer:

D. The sun's light helps plants grow

Explanation:

This is a scientific observation, because it is describing the structure of a living thing.

7 0
3 years ago
What is the mass of carbon in 290 g of CO2?
Elis [28]

Answer:

The mass of CO22 in total is 264 g.

The atomic mass for C is 12 g / mole.

The molar mass for CO22 is (12 + (2 × 16)) = 44 g / mole.

m C = (12 / 44) × 264 = 72 g

So, there are 72 g of C in 264 g of CO2

Explanation:

there's the answer have a good day.

4 0
3 years ago
4.65 L of nitrogen at standard pressure is compressed into a 0.480 L container. What is the new pressure in kPa?
Semenov [28]
Assuming that nitrogen gas is ideal, we can use the equation PV = nRT to relate first conditions to the second condition. At constant temperature, pressure and volume are indirectly related as follows:

P = k / V

k is equal nRT

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6 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
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