What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that
does its temperature vary by 25 ° C?
Answer:
143.75cal
Explanation:
Given parameters:
Mass of steel = 50g
Specific heat capacity of the steel = 0.115cal/g°C
Temperature = 25°C
Unknown:
Amount of heat = ?
Solution:
The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.
Amount of heat = m C (ΔT)
m is the mass
c is the specific heat capacity
ΔT is the temperature change
Now insert the parameters and solve;
Amount of heat = 50 x 0.115 x 25
Amount of heat = 143.75cal
1) Chemical equation
2Al + 6 HCl ---> 2Al Cl3 + 3 H2
2) molar ratios
2 mol Al : 3 moles H2
3) Proportion
2 mol Al / 3mol H2 = x / 9 mol H2
4) Solve for x
x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag
Answer: 6 moles
Answer:
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Explanation:
Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
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 = 7°C - 84°C
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj
Ammonium is NH₄⁺ and Carbonate is CO₃⁻² => Ammonium Carbonate is (NH₄)₂CO₃
