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eimsori [14]
3 years ago
8

The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa

ter, and steam are 2.09 j/g-k, 4.18 j/g-k, and 1.84 j/g-k, respectively. for h2o, hfus = 6.01 kj/mol, and hvap = 40.67 kj/mol
Chemistry
1 answer:
kogti [31]3 years ago
8 0
First, calculate for the amount of heat used up for increasing the temperature of ice.

      H = mcpdT
       H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
    H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
    H = (18 g)(4.18 J/gK)(70 k)
    H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J
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Answer:

Explanation:

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5 0
2 years ago
The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium
PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

 \frac{9 .84x10^8}{1000000}      = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = \frac{mass}{volume}

mass = density x volume

mass = 1 x 984

         = 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

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