42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where,
is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
=
×ΔT
where
= Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
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Answer:
This solution has a volume of 98.4 mL
Explanation:
Step 1: Data given
Molarity of AgClO4 solution = 1.27 mol/L
Number of moles AgClO4 = 125 mmol = 0.125 mol
Molar mass of AgClO4= 207.32 g/mol
Step 2: Calculate volume of the 1.27 M solution
Molarity = moles / volume
Volume = moles / molarity
Volume = 0.125 moles / 1.27 mol /L
Volume = 0.0984 L = 98.4 mL
This solution has a volume of 98.4 mL
Answer:
sulfite ion SO32
Explanation:
A molecular ion is a covalently bonded set of two or more atoms, or of a metal complex, that can be considered to behave as a single unit and that has a net charge that is not zero.
A answer.
" All wavelengths of visible light that shine on it. "
it will not reflect 100 percent of the brightness.
HOPE THIS HELPS
The answer is D solubility
hope this helps:)