Answer:
Option B.
Explanation:
As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.
Then 2 moles of ethane will produce 4 moles of CO₂
The correct unabbreviated electron configuration is as below
Vanadium - 1S2 2S2 2P6 3S2 3p6 3d3 4s2
Strontium - 1s2 2S2 2P6 3S2 3P6 3d10 4S2 4P6 4S2
Carbon =1S2 2S2 2P2
<u><em> Explanation</em></u>
vanadium is in atomic number 23 in the periodic table hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Strontium is in atomic number 38 in periodic table hence its electron configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4s2
Carbon is in atomic number 6 in periodic table therefore its electron configuration is 1s2 2s2 2p2
When a non metal gains or share by two electrons then the valency will be 2^- or 2^+
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
Answer:
25 + 273= 298k.
Explanation:
I guess it's correct answer
