Answer:
[H2]eq = 0.0129 M
[F2]eq = 1.0129 M
[HF]eq = 0.9871 M
Explanation:
∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2
experiment:
∴ n H2 = 3.00 mol
∴ n F2 = 6.00 mol
∴ V sln = 3.00 L
⇒ [H2]i = 3.00 mol / 3.00 L = 1 M
⇒ [F2]i = 6.00 mol / 3.00 L = 2 M
[ ]i change [ ]eq
H2 1 1 - x 1 - x
F2 2 2 - x 2 - x
HF - x x
⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2
⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115
⇒ x² = (2 - 3x + x²)(115)
⇒ x² = 230 - 345x + 115x²
⇒ 0 = 230 - 345x + 114x²
⇒ x = 0.9871
equilibrium:
⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M
⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M
⇒ [HF] = x = 0.9871 M
Answer:
0.74mol
Explanation:
4.5×10^23/6.02×10^23=0.74mol
Answer:
See explanation
Explanation:
According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.
Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.
For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.
Its B i believe, Storms and currents can bring in sediments from other places
The answer is NO. During combustion of hydrogen, water is produced as an end product. The chemical formulae for combustion of hydrogen are:
2H2 + O2 -->2 H2O
another advantage is that the whole of the fuel load is combusted hence little wastage. This has made the fuel advantageous in its use as rocket fuel since there is no baggage fuel.
