Thủy phân este trong môi trường kiềm, thu được natri axetat và ancol metylic. công thức X là gì?
1 answer:
You might be interested in
Answer:
.
Explanation:
Magnesium chloride and silver nitrate reacts at a ratio:
.
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
.
The precipitate silver chloride is insoluble in water and barely ionizes. Hence,
isn't rewritten as ions.
Net ionic equation:
.
Calculate the initial quantity of nitrate ions in the mixture.
.
Since nitrate ions do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
.
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be of the concentration in the original solution.
.
Answer:
a) molar composition of this gas on both a wet and a dry basis are
5.76 moles and 5.20 moles respectively.
Ratio of moles of water to the moles of dry gas =0.108 moles
b) Total air required = 68.51 kmoles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.
Explanation:
Let assume we have 100 g of mixture of gas:
Given that :
Mass of methane =75 g
Mass of ethane = 10 g
Mass of ethylene = 5 g
∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)
Their molar composition can be calculated as follows:
Molar mass of methane
Molar mass of ethane
Molar mass of ethylene
Molar mass of water
number of moles =
Their molar composition can be calculated as follows:
4.69 moles
0.33 moles
0.18 moles
0.56 moles
Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles
= 5.76 moles
Total moles of gas for dry basis = (5.76 - 0.56)moles
= 5.20 moles
Ratio of moles of water to the moles of dry gas =
=
= 0.108 moles
b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:
4.69 2× 4.69
moles moles
0.33 3.5 × 0.33
moles moles
0.18 3× 0.18
moles moles
Mass flow rate = 100 kg/h
Their Molar Flow rate is as follows;
Total moles of required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
= 11.075 k moles.
In 1 mole air = 0.21 moles
Thus, moles of air required =
= 52.7 k mole
30% excess air = 0.3 × 52.7 k moles
= 15.81 k moles
Total air required = (52.7 + 15.81 ) k moles/h
= 68.51 k moles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.