Answer:
if you are working with hazardous materials.
Explanation:
A properly operating and correctly used fume hood can reduce or eliminate exposure to volatile liquids, dusts, and mists. It is advisable to use a laboratory hood when working with all hazardous substances.
There is 6 non - bonding pairs.
Let me show you one easy method to do this.
o22-, oxygen valence electron = 6 here we have two so total 12, and -2 that means we add electrons so it’s all equal to 14 right.
whenever need to find lone pair, subtract the number you get with the lowest multiple of 8.
here we total 14 valence electron right so lowest multiple of 8 would be 8.
so 14 - 8 = 6 and that is our answer.
Let me know if you have Problem with chemistry.
Explanation:
No of molecules=0.500×6.023×10²³=3.011×10²³ molecules
The pH of pure water is neutral because the concentration of hydronium ions equals that of hydroxide ions.
Answer: Option A
<u>Explanation:</u>
Water is one of the most important constituents of living being. It is said that there is no life without water. So that water need to neutral in nature to save life. Pure water is composed of hydronium 
and hydroxide 
ions. It is known that hydronium ions are acidic in nature with concentration of 
.
Similarly, the hydroxide ions which are basic in nature will be in same concentration as that of hydronium ions. So, as the concentration of basic and acidic elements are equal with the same strength of pH, the combination of these ions lead to formation of pure water with the pH being neutral.
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
