When cooled by liquid nitrogen, the balloon shrinks (not as much as the air-filled balloon) and it sinks down on the table. When heating up, the balloon slowly rises and flies up in the air again. Explanation 1: The volume of the balloon decreases by the low temperature, because the gas inside is cooled down.
Answer:
0.062mol
Explanation:
Using ideal gas law as follows;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821Latm/molK)
T = temperature (K)
Based on the information provided;
P = 152 Kpa = 152/101 = 1.50atm
V = 0.97L
n = ?
T = 12°C = 12 + 273 = 285K
Using PV = nRT
n = PV/RT
n = (1.5 × 0.97) ÷ (0.0821 × 285)
n = 1.455 ÷ 23.39
n = 0.062mol
Evaporation happens<span> when atoms or </span>molecules<span> escape from the liquid and turn into a vapor. Not all of the </span>molecules in a liquid have the same energy. <span>Sometimes a </span>liquid<span> can be sitting in one place (maybe a puddle) and its molecules will become a </span>gas<span>. That's the process called </span>evaporation<span>. It can happen when liquids are cold or when they are warm. It happens more often with warmer liquids. You probably remember that when matter has a higher temperature, the molecules have a higher </span>energy<span>. When the energy in specific molecules reaches a certain level, they can have a </span>phase change<span>. Evaporation is all about the energy in individual molecules, not about the average energy of a system. The average energy can be low and the evaporation still continues. </span>
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
