A 44.8 g rhodium sample contains how many rhodium atoms
1 answer:
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Answer: 2.02 J/g°C
Explanation:
To find the heat capacity, we have to manipulate the equation for heat.
q=mCΔT becomes C=q/(mΔT) to find heat capacity. Since we are given our values, we can plug in to find C.
*Please ignore the capital A in front of the °C. In order to have ° in the equaiton, the A pops up.
Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L