In 0.190 mole of C6H14O, there is 0.190*6 (number of C in one molecule) = 1.140 mole of C atoms. The total number of C atoms = 1.14 * 6*
(atoms of C in one mole) = 6.84*
atoms.
Answer:
Mg(NO4)2 is 180.3 g/mol
Explanation:
First find the substance formula.
Magnesium Nitrate.
Magnesium is a +2 charge.
Nitrate is a -1 charge.
So to balance the chemical formula,
We need 1 magnesium atom for every nitrate atom.
2(1) + 1(-2) = 0
So the substance formula is Mg(NO4)2.
Now find the molar mass of Mg(NO4)2.
Mg = 24.3 amu
N = 14.0 amu
O = 16.0 amu
They are three nitrogen and twelve oxygen atoms.
So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol
So the molar is mass is 180.3 g/mol.
The final answer is Mg(NO4)2 is 180.3 g/mol
Hope it helped!
Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Answer:
suspension
Explanation:
a mixture in which particles can be seen and easily separated by settling or filtration.
hope this helps