1. 12.992 L
2. 2.42 moles
3. 275.52 L
4. 567.844 g
<h3>Further explanation</h3>
Given
moles and volume at STP
Required
mass, volume and moles
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
1. 0.58 moles ammonia :
Volume = 0.58 moles x 22.4 L = 12.992 L
2. 77.5 grams of O₂ :
Moles = 77.5 grams x (1 mol/32 grams) = 2.42
3. 12.3 mole of Bromine gas :
Volume = 12.3 mole x (22.4 L/1 mole) = 275.52 L
4. 4.8 moles iron(II)chloride :
Mass = 4.48 moles x molar mass ( 126,751 g/mol) = 567.844 g
4 protons , 3 protons , 2 protons
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Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT


If: ΔT (T final - T initial) = ?

