Answer:
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
Explanation:
Step 1: Data given
Temperature = constant
Volume will be doubled
Number of molecules will be doubles
Step 2:
p*V = n*R*T
⇒ gas constant and temperature are constant
Initial pressure = n*R*T / V
Initial pressure = 2*R*T/2
Initial pressure = RT
Final pressure = 4*RT / 4
Final pressure = R*T
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
Iron rusts when exposed to air → chemical property
Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.
b) they lose 2 electrons as they are transferred to the non metal.
c)They obtain this charge as when they are made into an ionic compound the 2 electrons on the outer shell are transferred to the non metal, meaning there are 2 more protons that electrons, giving it a positive charge.
hope this helps! :)
