Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: 
Molar mass of A = 
The depression in freezing point of solution with B solute: 
Molar mass of B = 
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
This means compound B has greater molar mass than compound A,
Given :
Number of molecules of hydrogen peroxide, N = 4.5 × 10²².
To Find :
The mass of given molecules of hydrogen peroxide.
Solution :
We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.
So, number of moles of hydrogen peroxide is :
Now, mass of hydrogen peroxide is given as :
m = n × M.M
m = 0.0747 × 34 grams
m = 2.54 grams
Hence, this is the required solution.
14 since K has 1 valence but there’s two so 2 valence for k and oxygen has 6 but there’s two so 12
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
So you need to put numbers before each compound to make sure there are the exact same number of elements on each side. If you put a 4 before NH4 there are 4 Nitrogen and now 16 hydrogen. I just played around with numbers and guessed until I got them even.
