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
Chemistry affects all aspects of life and most natural events because all living and nonliving things are made of matter.
Answer:
See explanation.
Explanation:
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In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:
To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.
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MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
Answer:
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Explanation:
1. 2)
2. 3)
3. 4) Sr
4. 3)
5. 4)
6. 2)
7. 1)
8. 4)
9. 3)
10. 3)
11. SO3, H2SO4, Na2SO4
12.
A) оксид меди (II) 2) CuO
Б) хлорид меди(II) 4) CuCl2
В) сульфит меди (II) 3) CuSO 3
Г) гидроксид меди (II) 1) Cu(OH)2
13.
1. Fe+HCl= б) FeCl 2 +H 2
2.Fe+O2= в) Fe 3 O 4
3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O
4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl
14. 2Ca + O2 = 2CaO
CaO + H2O = Ca(OH)2
Ca(OH)2 + 2HCl = CaCl2 + 2H2O
