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Hoochie [10]
3 years ago
8

What happens to the energy that is lost when water freezes?

Chemistry
2 answers:
vlada-n [284]3 years ago
5 0

When water freezes it gives up some of the water's energy. This energy that is given up is the latent heat of freezing. When the water was freezing latent heat of freezing energy was being released. ... When a latent heat process occurs the temperature remains constant.

SVEN [57.7K]3 years ago
4 0
<span>As we know through the principle of conservation of energy, energy can neither be created nor destroyed. Therefore, the energy removed from the water in order to make it freeze is absorbed by the surroundings. This is why the surroundings in which freezing is taking place are below freezing. This is more easily illustrated in the example of condensation. If you were to hold a plate over a pot of boiling water, some of the water would give its energy to the plate and condense on its surface.</span>
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A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give
8_murik_8 [283]

<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}

Where,

m_{solute} = Given mass of solute (KOH) = 3.301 g

M_{solute} = Molar mass of solute (KOH) = 56.1 g/mol

W_{solvent} = Mass of solvent = 96.699 g

Putting values in above equation, we get:

\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m

Hence, the molality of potassium hydroxide solution is 0.608 m

4 0
3 years ago
What is Nuclear fusion​
IRINA_888 [86]

Answer:

Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles (neutrons or protons).

3 0
2 years ago
Enter the electron configuration for the ion most likely formed by phosphorus. Express your answer in the order of orbital filli
Bingel [31]

Answer:

[Ne] 2s2 2p3

Explanation:

Phosphorus will most likely have an ion that will be 3- because it wants to have a full outer shell. Thus, the elctron configuration is: 1s2 2s2 2p6 3s2 3p3.

4 0
3 years ago
what is the molarity of a RbOH solution if 60.0 mL of the solution is neutralized by 52.8 mL of a 0.5M HCl solution (Hint: Ma x
Dahasolnce [82]
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume 
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted 
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL 
Mb = 0.44 M 
molarity of base - 0.44 M 

7 0
2 years ago
Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo
Arada [10]

Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

3 0
2 years ago
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