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laiz [17]
3 years ago
11

When you boil a pot of liquid water, the water turns to steam and rises in the air. This is an example of what.

Chemistry
2 answers:
Thepotemich [5.8K]3 years ago
5 0

Answer:

C: evaporation

Explanation:

this is a liquid to gas, so evaporation matches this scenario

hope this helps! Please give brainiest!!! ty

Georgia [21]3 years ago
4 0

Answer:

Hello There!!

Explanation:

The answer is C: evaporation because evaporation is the process in which a liquid boils then evaporates and lastly becomes a gas.

hope this helps,have a great day!!

~Pinky~

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Unit: Stoichiometry
Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0
3 years ago
A 15.75 g piece of iron absorbs 1097 joules of heat energy, and its temperature changes from 25°C to 177°C. Calculate the specif
vekshin1

The answer for the following problem is mentioned below.

Explanation:

Given:

mass of iron (m) = 15.75 grams

heat (q) = 1097 J

initial temperature (t_{1}) = 25°C

final temperature (t_{2}) = 177°C

To find:

specific heat (c)

We know;

 c = q ÷ mΔT

where;

c represents the specific heat

q represents the heat

m represents the mass

t represents the temperature

c = \frac{1097}{15.75 * (177-25)}

c = 0.45 J/kg°C

<u><em>Therefore the specific heat capacity of iron is 0.45 J/kg°C.</em></u>

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