Answer:
Q = 233.42 J
Explanation:
Given data:
Mass of lead = 175 g
Initial temperature = 125.0°C
Final temperature = 22.0°C
Specific heat capacity of lead = 0.01295 J/g.°C
Heat absorbed by water = ?
Solution:
Heat absorbed by water is actually the heat lost by the metal.
Thus, we will calculate the heat lost by metal.
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 = T2 - T1
ΔT = 22.0°C - 125.0°C
ΔT = -103°C
Q = 175 g × 0.01295 J/g.°C×-103°C
Q = -233.42 J
Heat absorbed by the water is 233.42 J.
The chemical equation represents the reaction describes is;
4NH3 + 5O2 = 4NO + 6H2O
Therefore 4 moles of NH3 reacts with 5 moles of O2.
1 mole of O2 (molar mass) = 2 * 16 = 32g.
5 moles of O2 = 5 * 32 = 160g
4 moles of NH3 = 4 (14 + 3*1) = 68g
Therefore, 68g of NH3 reacts with 160g of O2.
But, we have only 4.5 g of oxygen.
68g reacts with 160g
Xg reacts with 4.5
X = 68*4.5 / 160 = 1.9125g
Explanation:
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