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Fiesta28 [93]
3 years ago
5

The heat of fusion of water is 80 cal/ g. How much energy is needed to melt 0.05 kg of ice?

Chemistry
1 answer:
Alex Ar [27]3 years ago
7 0
80 cal/ g is same as 80 calories in each 1 gram. so,

80 cal= 1 gram

first, we need to convert the kg into grams. all you have to do is multiply by 1000 or move the decimal three times to the right. 

0.05 kg= 50.0 grams

50.0 g (80 cal/ 1 gram)= 4000 calories

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an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with
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Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.  

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

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Which refers to the chemical name for NO2?
Anna35 [415]

Answer:

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Explanation:

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The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure
AlekseyPX

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

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\frac{P_1}{T_1}=\frac{P_2}{T_2}

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T_1 = initial temperature = 27^oC=273+27=300K

T_2 = final temperature = ?

Now put all the given values in the above equation, we get:

\frac{198kPa}{300K}=\frac{225kPa}{T_2}

T_2=340.9K\approx 341K

Therefore, the temperature of the air in the tire is, 341 K

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