Answer:
D/CH3COOH
Explanation:
CH3COOH is a weak acid, it partially dissociates into its ions in an aqueous solution.
In 250 mL of volumetric flask add 0.975875 grams of 
and dissolve it in the 250 mL of water.
Given:
The solid of calcium fluoride.
To prepare:
The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.
Method:
Molarity of the fluoride ion solution needed = M = 0.100 M
The volume of the fluoride ion solution needed = V = 250 mL
The moles of fluoride ion needed = n
According to the definition of molarity:
Moles of fluoride ion = 0.025 mol
We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.
According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:
Moles of calcium fluoride = 0.0125 mol
Mass of calcium fluoride needed to prepare the solution :
Preparation:
- Weight 0.975875 grams of calcium fluoride
- Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
- Now add a small amount of water to dissolve the calcium fluoride completely.
- After this add more water up to the mark of the volumetric flask of volume 250 mL.
Learn more about molarity of solution ere:
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Answer:
F2(g)/HF(aq)>Co3 (aq)/Co2 (aq)> H2O2(aq)/H2O(l)> PbO2(s)/Pb(s)>Br2(l)/Br-(aq)
Explanation:
The tendency of any specie to function as oxidizing agent is a highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.
This implies that we could know a good oxidizing agent by looking at their respective reduction potentials. The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent. If there are a number of couples at having different reduction potentials, the order of oxidizing ability can be obtained by arranging the species in order of decreasing positive reduction potentials just as we have done in the answer above.
Answer:
0.67 L
Explanation:
Given data:
Volume of carbon dioxide = ?
Mass of nitroglycerine = 2.5 g
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
4C₃H₅N₃O₉ → 12CO₂ + 10 H₂O + 6N₂ + O₂
Number of moles of nitroglycerine:
Number of moles = mass/ molar mass
Number of moles = 2.4 g/ 227.1 g/mol
Number of moles = 0.01 mol
Now we will compare the moles of nitroglycerine and carbon dioxide from balance chemical equation.
C₃H₅N₃O₉ : CO₂
4 : 12
0.01 : 12/4×0.01 = 0.03 mol
Volume of CO₂:
PV = nRT
V = nRT/P
V = 0.03 mol× 0.0821 atm.L/mol.K× 273.15 K / 1 atm
V = 0.67 L
The answer is True statement.
