Molar mass:
KCl = 74.55 g/mol
KClO3 = 122. 55 g/mol
<span>Calculation of the mass of KClO3 :</span>
<span>2 KClO3 = 2 KCl + 3 O2</span>
2* 122.55 g KClO3 ------------------ 2 * 74.55 g KCl
mass KClO3 ?? --------------------- 25.6 g KCl
mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55
mass KClO3 = 6274.56 / 149.1
mass = 42.082 g of KClO3
Therefore:
1 mole KClO3 ---------------------- 122.55 g
?? moles KClO3 ------------------- 42.082 g
moles KClO3 = 42.082 * 1 / 122.55
moles KClO3 = 42.082 / 122.55
=> 0.343 moles of KClO3
Answer C
hope this helps!
solution:
1000 = m*2400*(78-22) + m*8.79*10^5
1000= 134400m + 879000m
1000= 1030200m
m = 1000/1013400
m= 1013.4 grams
the final answer is 0.9706 grams
The solutions vapor pressure would be lower.
Ammonia is colorless gas with a characteristic smell. Its density is 0.589 times than air which makes it lighter than air. Ammonia can be easily liquefied due to the hydrogen bonding between the molecules. The boiling point is at -33.3 degrees Celsius and the freezing point is at -77.7 degrees Celsius.
Answer;
= 0.054 kg or 54 g
Explanation;
Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.
ΔT = 75 - 20 = 55 C.
Solve the equation for m
m = Q/ cΔT
Mass = 12500 / (55 × 4200)
= 0.054 kg or 54 g
