Answer:
2.5 L
Explanation:
The stoichiometric ratio of N2 to N2O4 is 1:1
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
Answer: C
Explanation:
Indicators are organic weak acids or bases with complicated structures.
