Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
Answer:
The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.
Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.
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
Answer:
BaI2
Explanation:
Hello, since the electronegativity of Barium and Iodine are 0.89 and 2.66, respectively, the difference is 1.77, so the bond is ionic.
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