Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j
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Many homeowners treat their lawns with CaCO3(s) to reduce
the acidity of the soil. The net ionic equation for the reaction of CaCO3(s)
with a strong acid, HCl (I chose HCl because it is a strong acid) is CaCO3(s) +2
HCl(aq) → CaCl2(s) + H2O(aq) + CO2(g).
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
