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liq [111]
3 years ago
5

How much heat, in kJ, is required to melt 54.0 g of ice at 0°C into water at 0°C if Hfus for water = 6.01?

Chemistry
2 answers:
FrozenT [24]3 years ago
7 0

18 that is what i think it is

kati45 [8]3 years ago
3 0
The heat of fusion of 1 mole of water is 6.01 kJ/mol
1 mole of water is equal to 18 g
Therefore; 54,0 g of water is equivalent to 54/18 = 3 moles of water
Hence heat required;
 = 3 × 6.01 = 18.03 kJ
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What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a
Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

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Answer:

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Answer:

hope this helps

Explanation:

glycosidic bond

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Answer:

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c=λν

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