This example is an controversial matter because melting an candle is considered as both physical change and chemical change because no new substances were made by melting an candle. The final answer is C.
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:
A white precipitate of magnesium hydroxide is formed in the above reaction.
Ionic form of the above equation follows:
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
Balanced equation have equal number of atoms of different elements in the side of reactants and products.
actual yield of ethanol = 305.0g
molar mass of sucrose = 342g
molar mass of ethanol =46g
mass of sucrose = 665g
mole of sucrose = mass / molar mass = 665/342
mole of sucrose =1.9 mole
sucrose : C2H5OH
1 : 4
1.9 : 1.9x4 =7.6 mole of C2H5OH are formed
mass (therotical yield ) of C2H5OH= mole x mass
mass (therotical yield ) of C2H5OH= 7.6 x 46 = 349.6g
percent yields of ethanol = actual /therotical x100
=305/349.6x100 = 87.24 %
