Answer:
A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
Explanation:
Hello,
A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
Best regards.
Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
Hahahaaaa none of the above but IF <span>(c) is
1/2 mole of NaCl and 1/3 mole of MgCl2 instead,
then C is the right ans :)</span>
Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;
The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A 
The density of sample B 
Therefore, sample B contains the larger density
