Answer:
B and D could be true
Explanation:
A volume of sodium hydroxide less than expected could occurs for two reasons:
The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:
A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration. FALSE. A long expose to the air decreases concentration of the NaOH.
B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use. TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.
C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use. FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.
D. The pipette used to deliver the vinegar solution was rinsed with water prior to use. TRUE. Again, you are adding a less amount of Vinegar than expected doing the necessary NaOH during titration less than expected
The class of compounds that ammonia belongs to is it is categorized as a basic compound. As it posses a pH value or rating above 7, which is characteristic of basic substances, solutions.
Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
It would be: 1s2, 2s2, 2p4
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles
The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
