How much heat is lost when solid Al with a mass of 4110 g cools from 660.0 oC to 25.0 oC?
1 answer:
Answer:
2,354,084.7 J
Explanation:
-We apply the Specific Heat Capacity formula to solve for the heat loss:
where
m is the mass of the substance
Q is the heat loss/gain
c is the substance's specific heat
T is the change in temperature
\Substitute given values in the equation to solve for Q:
Hence, the amount of heat lost is 2,354,084.7 J
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Answer:
M
Explanation:
The concentration of the analyte in the 5-mL flask would be M
This is a problem of simple dilution that can be solved using the dilution equation;
C1V1 = C2V2,
where C1 = initial concentration, V1 = initial volume, C2 = final concentration, and V2 = final volume.
<em>In this case, the initial concentration (C1) is not known, the initial volume (V1) is 1.00 mL, the final concentration is 6.97 x 10-5 M, and the final volume is 10.00 mL.</em>
Now, let us make the initial concentration the subject of the formula from the equation above;
C1 = C2V2/V1. Solve for C1 by substituting the other parameters.
C1 = 6.97 x 10-5 x 10/1 = M