Answer:
1-Chloropropane is likely the answer (attached a picture)
Explanation:
First off there are 3 peaks and 3 carbons which indicates to me that this will be a chain without any symmetry and that each carbon has hydrogens on it.
Second the triplet at 1.0 that integrates to 3 likely correlates to a CH3 (methyl) group. Peaks are very upfield triplets that integrate to three are almost always methyl peaks.
Third the triplet at 3.7 is indicative of being next to the halogen. Hydrocarbons by themselves do not have peaks that far downfield meaning that its shift could only be explained by the chlorine being involved. Also we know that this can't be next to the methyl group since its multiplicity is to low to be next to it.
That leaves the multiplet at 1.75 being the hydrogens on the middle carbon which also makes sense since it is more downfield then the methyl group (due to being closer to the chlorine) but is not far enough downfield to say the chlorine is there. It also makes sense that it is a multiple since it would be a hextet due to the adjacent 5 hydrogens which can't always be resolved.
I hope this helps and let me know if anything is unclear or needs further explanation.
Answer:
Explanation:
Given that:
From above:
To predict the effect of the addition of Br₂(g);
The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr
The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.
Answer : The value of and K is, -180 kJ/mol and
Explanation :
The balanced cell reaction will be,
The half-cell reactions are:
Oxidation reaction (anode) :
Reduction reaction (cathode) :
Relationship between standard Gibbs free energy and standard electrode potential follows:
where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:
Now we have to calculate the value of 'K'.
where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant =
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:
Therefore, the value of and K is, -180 kJ/mol and