Atomic <span>mass He = 4.002 u.m.a
4.002 g --------------- 6.02x10</span>²³ atoms
50 g ------------------ ? atoms
50 x ( 6.02x10²³) / 4.002
= 3.01x10²⁵ / 4.002
= 7.52x10²⁴ atoms of He
I believe its 10^23 u r saying
then no. of moles= 9.03*10^23/6.02*10^23=1.5mole
We are given the amount of Nitrogen gas and hydrogen gas reacted to form ammonia:
N2 = 19.25 grams
H2 = 11.35 grams
Set-up a balanced chemical equation:
N2 + 3H2 ==> 2NH3
The theoretical amount of ammonia that will be produced from the given amounts is:
First, we need to determine the limiting reactant to serve as our basis for calculation.
number of moles / stoichiometric ratio
N2 = 19.25 g/ 28 g/mol / 1 = 0.6875
H2 = 11.35 g/ 2 g/mol /3 = 1.89
The limiting reactant is N2.
0.6875 moles N2 * (2 NH3/ 1 N2) * 17 g/mol NH3
The amount of NH3 produced is 23.375 grams of ammonia. <span />
1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
</span>
AX(aq) is soluble and <span>BY(aq) is insoluble
the answer is
</span><span>E. BY</span>
