about 43001 is it i think
Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.
Explanation:
Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.

As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.
According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.
Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.

As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.

Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.
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.
Answer:
m = 671 grams
Explanation:
Given that,
No of moles, n = 4.9
Molar mass of Barium, M = 137 g
Mass divided by molar mass is equal to no of moles. It can be given by the formula as follows :

or
m = 671 grams
So, the total mass of the sample of Barium is 671 grams.