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lara [203]
3 years ago
6

Energy increases heat and causes a phase change from solid to liquid or liquid to solid

Chemistry
1 answer:
weqwewe [10]3 years ago
7 0
<span>So, how could there be a change in heat during a state change without a change in temperature? During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same.</span>

 

Below is a picture of solid ice melting into liquid water. The molecule of ice and the molecule of water (the black balls) are moving with the same rate of vibration in this diagram. This is meant to show that they have the same average speed and thus the same average kinetic energy (since they have the same mass) and thus the same Kelvin temperature. The motions are, though, greatly exaggerated. Actually, the motions of the molecules should be considered tiny vibrations.

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Dew forming on grass is an example of condensation true or fales and explain
exis [7]
True becuase dew is coming out of the air which if you look at a glass of water it has condensation on it becuase it is hot
7 0
4 years ago
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A drought is best defined as
Lapatulllka [165]
A drought is a period of drier-than-normal conditions that results in water-related problems.<span> When rainfall is less than normal for several weeks, months, or years, the flow of streams and rivers declines, water levels in lakes and reservoirs fall, and the depth to water in wells increases.</span>
5 0
3 years ago
If 20.0 g LIOH react with excess KCl, giving a 17.0
adell [148]

The yield of lithium chloride is 1.92 grams.

Option D.

<h3><u>Explanation:</u></h3>

In this reaction, we can see that 1 mole of lithium hydroxide reacts with 1 mole of potassium chloride to produce 1 mole of lithium chloride and 1 mole of potassium hydroxide.

Molecular weight of lithium hydroxide is 24.

Molecular weight of lithium chloride is 42.5.

So 24 grams of lithium hydroxide produces 42.5 grams of lithium chloride.

So, 20 grams of lithium hydroxide produces 20 * 24 / 42.5  grams =11. 29 grams of lithium chloride.

But this is when the yield is 100%.

But yield is 17%.

So the yield is 1.92 grams of lithium chloride.

8 0
3 years ago
Calculate the volume of a 3.50 M solution of H2SO4 made from 49 g of H2SO4
maxonik [38]

Answer:

0.143L

Explanation:

Molar mass of H2SO4 = 98g/Mol

No of mole = mass/molar mass

No of mole= 49/98 = 0.5 mol

No of mol = concentration × volume

Volume = n/C = 0.5/3.5 = 0.143L

3 0
3 years ago
1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
3 years ago
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