H2(g) +C2H4(g)→C2H6(g)
H-H +H2C =CH2→H3C-Ch3
2C -H bonds and one C-C bond are formed while enthalpy change (dH) of the reaction,
H-H: 432kJ/mol
C=C: 614kJ/mol
C-C: 413 kJ/mol
C-C: 347 kJ/mol
dH is equal to sum of the energies released during the formation of new bonds or negative sign, and sum of energies required to break old bonds or positive sign.
The bond which breaks energy is positive.
432+614 =1046kJ/mol
Formation of bond energy is negative
2(413) + 347 = 1173 kJ/mol
dH reaction is -1173 + 1046 =-127kJ/mol
Answer:
Explanation:
It can be determined by measuring the Ph. D is incorrect.
C: is wrong because if you are making something acidic, you are increasing the H+
B: is the correct answer.
A: pH decreases. H+ increases which makes the Ph decrease. It is an oddity of the formula that makes this happen.
Answer:- oxygen.
Explanations:- The electronic configuration is given and we are asked to figure out the electrically neutral atom that will have the electron configuration, 
.
The sum of electrons for this electron configuration is 8. If we look at the periodic table then 8 is the atomic number of oxygen.
So, the electrically neutral atom for the given electron configuration is oxygen.
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
Answer:
The calcium concentration must be greater outside the cell than inside the cell.
Explanation:
My previous answer was deleted from the explanation I provided from another website.
