Answer:
The concentration of acetic acid in vinegar of that trial would be <u><em>greater than</em></u> the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
Answer:
0.258 mg of iron remains.
Explanation:
To solve this problem we can use the formula
M₂ = M₀ * 
Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.
Using the data given by the problem:
M₂ = 2.000 mg * 
M₂ = 0.258 mg
11.43g/mL
Explanation:
Given parameters:
Volume of water in the graduated cylinder = 100mL
Volume of water + lead weight = 450mL
Mass of lead weight = 4000g
Unknown:
Density of the lead weight = ?
Solution:
Density is the mass per unit volume of a body.
Density = 
Volume of the lead weight = volume of water displaced
Volume of lead weight = 450 - 100 = 350mL
Density =
= 11.43g/mL
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Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g