Remember: heat lost = heat gained
When calculating heat loss or gain, remember
mass*(spec heat cap)*(change in T)
The unknown loses heat- we don't know the spec heat cap, so we'll call it x.
The water gains. I've omitted the units, but always use when solving problems on your own.
75*x*(96.5-37.1) = 1150*4.184*(37.1-25)
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC
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Answer is: D. It is not sodium bicarbonate.
Balanced chemical reaction of heating sodium bicarbonate: 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.
This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.
<span>128 g/mol
Using Graham's law of effusion we have the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = rate of effusion for gas 1
r2 = rate of effusion for gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Since the atomic weight of oxygen is 15.999, the molar mass for O2 = 2 * 15.999 = 31.998
Now let's subsitute the known values into Graham's equation and solve for m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
127.992 = m2
So the molar mass of the unknown gas is 127.992 g/mol.
Rounding to 3 significant figures gives 128 g/mol</span>
