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Gnesinka [82]
3 years ago
9

Compare and contrast 10kg of melting ice and 1kg of freezing water address temperature heat flow thermal energy what is the simp

lified answer to that.
Chemistry
1 answer:
iris [78.8K]3 years ago
7 0

Answer:

10 kg of ice will require more energy than the released when 1 kg of water is frozen because the heat of phase transition increases as the mass increases.

Explanation:

Hello!

In this case, since the melting phase transition occurs when the solid goes to liquid and the freezing one when the liquid goes to solid, we can infer that melting is a process which requires energy to separate the molecules and freezing is a process that releases energy to gather the molecules.

Moreover, since the required energy to melt 1 g of ice is 334 J and the released energy when 1 g of water is frozen to ice is the same 334 J, if we want to melt 10 kg of ice, a higher amount of energy well be required in comparison to the released energy when 1 kg of water freezes, which is about 334000 J for the melting of those 10 kg of ice and only 334 J for the freezing of that 1 kg of water.

Best regards!

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A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?
Pachacha [2.7K]
<span>The student should follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
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1. First he should calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>

</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L

Molarity = number of moles / volume of the solution

Hence, number of moles in 1 L = 2 mol

2. Find out the mass of dry CaCl</span>₂ in 2 moles.<span>

moles = mass / molar mass

Moles of CaCl₂ = 2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol

Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
                                     = 221.96 g

3. Weigh the mass accurately 

4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and finally wash the funnel and watch glass with de-ionized water. That water also should be added into the volumetric flask.

5. Then add some de-ionized water into the volumetric flask and swirl well until all salt are dissolved.

<span>6. Then top up to mark of the volumetric flask carefully. 
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3 years ago
If 30.5 g of 15% potassium nitrate solution are reacted to excess magnesium chloride how many grams of potassium chloride will b
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Here the step by step and answer

4 0
2 years ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{55\ g}{26.981539\ g/mol}

Moles= 2.0384\ mol

According to the reaction:-

4Al+3O_2\rightarrow 2Al_2O_3

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with \frac{3}{4} moles of oxygen gas

2.0384 moles of aluminum react with \frac{3}{4}\times 2.0384 moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

7 0
3 years ago
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