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Nuetrik [128]
2 years ago
10

HELP PLEASE I HAVE A TEST TODAY AND I DON'T UNDERSTAND ANY OF THIS...

Chemistry
1 answer:
myrzilka [38]2 years ago
5 0

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

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Explanation:

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3 years ago
If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
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Answer:

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For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

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\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

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3 years ago
The specific heat of zinc is 0.39 J/g*°C. How much energy needed to change the temperature of 34g of zinc from 22°C to 57°C. Is
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Answer:

464.1 J absorbed.

Explanation:

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