Explanation:
so moles = mass ÷ mr (1+ 79.9)
so 10.00g ÷ 80.9
which is 0.1236093943
so to 3 S.F is 0.124 moles
also there is 1 to 1 ratio for LiOH to HBr
hope this helps :)
Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
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
hydrogen and oxygen atoms
Explanation:
this is because they are non metals and there will be sharing of electrons between the two atoms forming the bond
Sodium (Na) has a +1 charge and Iodine ( I ) has a -1 charge. To create a molecule of sodium iodide the charges will need to balance.
Because the charges on anion and cation are the same; the molecular formula will be NaI
