Answer:
there are no valence electrons left over, so the molecule has four bond pairs and no lone pairs.
Explanation:
Answer:
A. 0.35 M
Explanation:
Hello,
In this case, given the volume and concentration of lithium hydroxide and the volume of chloric acid, we can compute the concentration of the neutralized acid by using the following equation:
Therefore, answer is A. 0.35 M.
Regards.
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
Answer:
24.9%
Explanation:
According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:
0.0927 × 100 = 9.27%
Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;
100% - 9.27% = 90.73%
THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles
N.B: mole = mass/molar mass
Given the Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
For NaCl;
0.0927 = mass/58.44
mass = 0.0927 × 58.44
5.42g
For H2O;
9.073 = mass/18.016
mass = 9.073 × 18.016
= 16.35g
Total mass of solution = 16.35g + 5.42g = 21.77g
Mass percent of NaCl = mass of NaCl/total mass × 100
% mass of NaCl = 5.42g/21.77g × 100
= 0.249 × 100
= 24.9%
