Answer:
when there is a radical in the denominator, we should rationalize (mutiply the denominator and numerator by the radical) to get rid of the radical in the denominator.
CH4+(x)O2=CO2 +(Y)H2O
C=1 +H=4 +O=? = C=1 +O=2+? +H=?
H=4>>Y=2
C=1 +H=4 +O=? = C=1 +O=(2+2) +H=4
C=1 +H=4 +O=4 = C=1 +O=4 +H=4
O=4>>X=2
CH4+(2)O2 =CO2 +(2)H2O
C:H:O = 40/12 : 6,7/1 : 53,3/16 = 3,33 : 6,7 : 3,33 ≈ 1 : 2 : 1
CH₂O
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
Answer:
4) 0.26 atm
Explanation:
In the process:
Benzene(l) → Benzene(g)
ΔG° for this process is:
ΔG° = -RT ln Q
<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>
ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm
1.336 = ln P(benzene) / 1atm
0.26atm = P(benzene)
Right answer is:
<h3>4) 0.26 atm
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