1 mole of nitric acid produce 1 mole of ammonium nitrate.
moles in 5000 kg of ammonium nitrate :
( molecular mass of ammonium nitrate is 80 gm/mol )
So, number of moles of nitric acid required are also 62500 moles.
Mass of 62500 moles of nitric acid :
Hence, this is the required solution.
Answer:
Explanation:
You have to calculate cuz you can`t count that much and he volume is 999.0031.
Answer:
I think the answer is A.
please give thanks if it helps
and sorry if it doesn't help.
29.0 mL in liters:
29.0 / 1000 => 0.029 L
n = M x V
n = 0.290 x 0.029
n = 0.00841 moles of NaOH
hope this helps!
