Answer:
58.92 g EDTA
Explanation:
315.1 mL = .3151 L
M = Moles / Liter
.3151 L x <u>0.5 mol EDTA</u> x <u>374 g EDTA</u> = 58.92 g EDTA
1 L EDTA 1 mol EDTA
Answer:
See explaination
Explanation:
Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,
AgF + NaCl= AgCl + NaF
Here, the color of AgCl is white, hence the solution cannot be AgCl.
Determination of NaCl
Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.
Reactions
Ag+ (aq)+ Cl-(aq) = AgCl(aq)
Ag+(aq) + SCN-(aq) = AgSCN(aq)
Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)
Answer:
Explanation:
<em>Telophase is the process that separates the duplicated genetic material carried in the nucleus of a parent cell into two identical daughter cells.</em>
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Answer:
V = 34430 mL
Explanation:
Given data:
Volume in mL = ?
Number of moles of gas = 2.00 mol
Temperature = 36°C (36+273= 309K)
Pressure of gas = 1120 torr
Solution:
Formula:
PV = nRT
V = nRT/P
V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr
V = 38563.2 torr • L / 1120 torr
V = 34.43 L
L to mL
34.43 L ×1000 mL / 1 L
34430 mL
The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g
The equation is Mr=Mi(1/2)^n
where n is the number of half-lives
Mr is the mass remaining after n half lives
Mi is the initial mass of the sample
To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)
n=1000/24100=0.0414
So Mr=5x(1/2)^1=2.5 g
The mass remaining is 2.5 g
- The half life is the time in which the concentration of a substance decreases to half of the initial value.
Learn more about half life at:
brainly.com/question/24710827
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