Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.
We make a proportion out of the word problem
(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
Answer:Nuclear binding energy is the energy needed to separate nuclear particles
The strong nuclear force holds an atom’s protons and neutrons together
Nuclear binding energy can be calculated using E=mc2
Explanation:
So the unbalanced equation would be Mg + N^2 --> Mg^3N^2
Which means the balanced equation would be 3Mg + N^2 --> Mg^3N^2
This is balance the equation out since you now has 3 magnesium and 2 nitrogen on the left side, and 3 magnesium on 2 nitrogen on the right. Double check my work though, it's been awhile.
