A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur
e is changed to 22.5oC, and the pressure is changed to 1.05 atm?
Please answer ASAP
Please answer ASAP
1 answer:
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Answer:
Explanation:
Given the moles, we are asked to find the mass of a sample.
We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.
Multiply by the given number of moles, which is 2.0
The moles of methanol will cancel each other out.
The denominator of 1 can be ignored.
Multiply.
There are 64 grams of methanol in the sample.
Answer:
The correct answer is option C.
Explanation:
1.0 g sample of a cashew :
Heat released on combustion of 1.0 gram of cashew = -Q
We have mass of water = m = 1000 g
Specific heat of water = c = 4.184 J/g°C
ΔT = 30°C - 25°C = 5°C
Heat absorbed by the water : Q
Heat released on combustion of 1.0 gram of cashew is -20,920 J.
3.0 g sample of a marshmallows :
Heat released on combustion of 3.0 g sample of a marshmallows = -Q'
We have mass of water = m = 2000 g
Specific heat of water = c = 4.184 J/g°C
ΔT = 30°C - 25°C = 5°C
Heat absorbed by the water : Q'
Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J
Heat released on 1.0 g sample of a marshmallows : q
Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.
-20,920 J. > -13,946.67 J
The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.