Explanation:
The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)
The mass of aluminum present in the pot is:
Hence, in the given pot 270g Al is present.
The gram atomic mass of Al -27 g/mol
Given the mass of Al is 270 g
Substitute these values in the above formula:
Answer is 10.0 mol of Al is present.
I think that different liquids have different freezing points because every liquid consists of different atoms and different things that make up the atom causing them to have different freezing points.
Answer:
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Explanation:
Answer:
T2 = 135.1°C
Explanation:
Given data:
Mass of water = 96 g
Initial temperature = 113°C
Final temperature = ?
Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)
Specific heat capacity of aluminium = 0.897 j/g.°C
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 = T2 - T1
Now we will put the values in formula.
Q = m.c. ΔT
1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C
1900 j = 86.112 j/°C × T2 - 113°C
1900 j / 86.112 j/°C = T2 - 113°C
22.1°C + 113°C = T2
T2 = 135.1°C
