Answer and Explanation:
a. The equation of K of this reaction is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
b. The moles of compound F is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
2 moles
Now, the mole of produced is
Now, we will the value of c by using the above equation
After solving the above equation we will get
0.5 moles
Answer: 1
Explanation: Step by Step
I'm pretty sure the answer is hypothesis
Answer: Thus molarity of is 1.35 M
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Thus molarity of is 1.35 M
<u>Answer:</u> The molar concentration of ethylenediminediacetic-dihydrate and sodium ions in solution is 0.1976 M and 0.3952 M respectively.
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (disodium ethylenediaminetetraacetic acid dihydate) = 36.7845 g
Molar mass of disodium ethylenediaminetetraacetic acid dihydate = 372.24 g/mol
Volume of solution = 0.5000 L
Putting values in above equation, we get:
As, 1 mole of disodium ethylenediaminetetraacetic acid dihydate produces 2 moles of sodium ion and 1 mole of ethylenediminediacetic-dihydrate.
Concentration of ethylenediminediacetic-dihydrate in solution =
Concentration of sodium ions in solution =
Hence, the molar concentration of ethylenediminediacetic-dihydrate and sodium ions in solution is 0.1976 M and 0.3952 M respectively.