Answer:
The correct option is;
A) The water molecules group together in clumps.
Explanation:
To answer the question, we look at each option as follows
A) The water molecules group together in clumps
This does not aid dissolution of compounds as clumping excludes non polar molecules, making them less soluble.
B) The partial negative oxygen atom end of the water molecule surround cations.
When an ionic compound such as NaCl is dissolved in water, the oxygen atoms surround the positive Na⁺ cations in the solution.
C) The partial positive hydrogen atom end of the water molecule surround anions
When the ionic NaCl compound is dissolved in water, the hydrogen atoms surround the negative Cl⁻ anions in the solution.
D) The water molecules also dissolve nonelectrolytes by creating a hydration sphere around the molecule.
Water is able to dissolve nonelectrolytes by creating an hydration sphere or hydration shell around the molecule.
Answer:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
