Answer:
Y³⁻ A non metal that gained three electrons.
V²⁺ A metal that lost two electron
Z²⁻ A non metal that gained two electrons
X³⁺ A metal that lost three electrons
Explanation:
Metals lost the electrons and form cation
while non metals gain the electrons and form anion.
Y³⁻
c) A non metal that gained three electrons.
Non metal gain three electrons and form anion with the charge of -3.
V²⁺
a) A metal that lost two electron
A metal lost two electrons and form cation with charge of +2.
Z²⁻
d) A non metal that gained two electrons
A non metal that gained two electrons and form anion with charge of -2.
X³⁺
b) A metal that lost three electrons
A metal that lost three electrons and form cation with charge of +3.
Answer: Special care is required to prepare a solution of sodium hydroxide or NaOH in water because considerable heat is liberated by the exothermic reaction. The solution may splatter or boil. Here is how to make a sodium hydroxide solution safely, along with recipes for several common concentrations of NaOH solution.
Explanation:
Hope this helps!
First step: calculate the number of moles of sodium reacted
moles=mass/molar mass
31.5g/23g/mol =1.37moles
from the reaction above the reacting ratio of Na to H2 is 2:1 therefore moles of H2 is 1.37/2=0.685
At STP 1 mole = 22.4L what about 0.685 moles
22.4 x0.685=15.344L
Answer:
the solubility of the ionic solid decreases
Explanation:
If a salt MX is added to an aqueous solution containing the solute AX, the X^- ion is common to both of the salts. The presence of X^- in the solution will suppress the dissolution of AX compared to the solubility of AX in pure water. This observation is known as common ion effect in chemistry.
The origin of common ion effect is based on Le Chatelier's principle. The addition of a solute will drive the equilibrium position towards the left hand side.
There number of calories that are required to raise the temperature of 105 g of water from 30.0 c to 70.0 c is calculated as follows
Q(heat)= mass(m) x specific heat capacity of water ( 1 cal/g/c) x delta T ( change in temperature
delta T = 70.0- 30.0= 40.0 c
Q= 105 g x 1 cal/g/c x 40 c = 4200 calories is needed to raise temperature of water from 30 c to 70 c