A. 1/2
Explanation- There is a 5/10 chance of choosing on of the numbers which simplifies to 1/2
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:
Several principal emissions result from coal combustion: Sulfur dioxide (SO2), which contributes to acid rain and respiratory illnesses.
Explanation:
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A weirdly worded question. That chemical formula is of glucose which is a carbohydrate as it consists of carbon, oxygen and hydrogen atoms and is a monosaccharide (simple sugar). It had 6 carbon atoms
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
