Answer:
0.17 moles
Explanation:
In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.
In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:
- Molar Mass of Ni (Nickel): 58.69 g/mol
- Molar Mass of C (Carbon): 12.01 g/mol
Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.
- 58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12
There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:
- 12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
- The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.
<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>
Answer:
16.05 amu
Explanation:
12.011 rounds to about 12.01 and 1.008 rounds to about 1.01 so when adding you'd do [12.01 + (1.01×4)]= 16.05
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Dissociation
NH₄NO₃(aq) = NH₄⁺(aq) + NO₃⁻(aq)
the hydrolysis of the cation
NH₄⁺(aq) + H₂O(l) = NH₃(aq) + H₃O⁺(aq)
pH<span><7</span>
