what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
1 answer:
Answer : The specific heat of the substance is 0.0936 J/g °C
Explanation :
The amount of heat Q can be calculated using following formula.
Where Q is the amount of heat required = 300 J
m is the mass of the substance = 267 g
ΔT is the change in temperature = 12°C
C is the specific heat of the substance.
We want to solve for C, so the equation for Q is modified as follows.
Let us plug in the values in above equation.
C = 0.0936 J/g °C
The specific heat of the substance is 0.0936 J/g°C
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Answer:
(B) 0.160 M
Explanation:
Considering:
Or,
Given :
For :
Molarity = 0.200 M
Volume = 20.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 20.0×10⁻³ L
Thus, moles of :
Moles of = 0.004 moles
For :
Molarity = 0.400 M
Volume = 30.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume =30.0×10⁻³ L
Thus, moles of :
Moles of = 0.012 moles
According to the given reaction:
1 mole of potassium carbonate react with 1 mole of barium nitrate
0.004 moles potassium carbonate react with 0.004 mole of barium nitrate
Moles of barium nitrate = 0.004 moles
Available moles of barium nitrate = 0.012 moles
Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of potassium carbonate gives 1 mole of barium carbonate
Also,
0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate
Mole of barium carbonate = 0.004 moles
Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)
Left over moles = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L
So, Concentration = 0.008/0.05 M = 0.160 M
<u>(B) is correct.</u>