M=D*V,
V=m/D
m=15 g
D=3 g/ml
V=15 g/3 g/ml=5 ml
18. Reaction will occur.
19. Reaction Will occur.
20. Reaction will occur.
21. Reaction will occur.
22. Reaction won't occur.
23. Reaction will occur.
24. Reaction will occur.
25. Reaction won't occur.
<h3><u>Explanation</u>:</h3>
The reaction rate of the metals with water, steam, acid, or hydroxides or their inert behavior towards them are noted in the metal activity series.
It contains all the metals one after the other which and the upper metal can replace the lower metal from its salt.
Calcium can replace hydrogen from acid, so the reaction will occur in 18. The products formed are calcium phosphate and hydrogen gas.
Chlorine is more reactive than bromine. So it can replace bromine from its salt to from bromine gas and magnesium chloride.
Aluminium can replace iron from its salt. So it will form aluminium oxide and iron metal. This reaction is used to obtain iron from ores.
Zinc can replace hydrogen from acid. So the products will be zinc chloride and hydrogen gas.
Chromium cannot displace hydrogen form water. So the reaction won't occur.
Tin can replace hydrogen form acid. So the reaction will proceed.
Magnesium will replace platinum from its salt. So magnesium oxide and platinum will form.
Bismuth cannot replace hydrogen from acid. So the reaction won't proceed.
Answer:
H₂SO₄ (aq) + 2LiOH (aq) ⇒ Li₂SO₄ (aq) + 2H₂O (l)
Explanation:
This is an acid-base reaction, so we know the products are going to a salt/ionic compound and water.
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.
