<u>Given:</u>
Initial volume of He, V1 = 19.2 L
Initial mass of He, m1 = 0.0860 g
Mass of He removed = 0.205 g
<u>To determine:</u>
The new volume of He i.e V2
<u>Explanation:</u>
Based on Avogadro's law:
Volume of a gas is directly proportional to the # moles of the gas
Volume (V) α moles (n) -----(1)
Atomic mass of He = 4 g/mol
Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles
Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles
Based on eq(1) we have:
V1/V2 = n1/n2
V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L
Ans: New volume is 14.6 L
Answer:
0.33 mol
Explanation:
Given data:
Volume of balloon = 8.3 L
Temperature = 36°C
Pressure = 751 torr
Number of moles of hydrogen = ?
Solution:
Temperature = 36°C (27 +273 = 300 K)
Pressure = 751 torr (751/760= 0.988 atm)
Formula:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
PV = nRT
0.988 atm × 8.3 L = n × 0.0821 atm.L/ mol.K ×
300 K
8.2 atm.L = n × 24.63 atm.L/ mol
n = 8.2 atm.L / 24.63 atm.L/ mol
n = 0.33 mol
First, we must find the total mass of CuBr₂:
Mass = 64 + 2 x 80
Mass = 224
Percentage mass of copper = mass of copper x 100 / total mass
Percentage mass of copper = (64 / 224) x 100
Percentage mass of copper = 28.45%
Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
