Answer:
1. The products of this reaction are ZnCl₂ and H₃PO₄.
2. 14.57 g.
Explanation:
<em>1. What would the products of this reaction be?</em>
- The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:
<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>
It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.
So, the products of this reaction are ZnCl₂ and H₃PO₄.
<em>2. If we produced 13.05 g of H₃PO₄, how many grams of hydrochloric acid would be need to start with?</em>
- Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:
n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.
<u><em>Using cross-multiplication:</em></u>
6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.
??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.
∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.
∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.
<em>So, the grams of hydrochloric acid would be need to start with = 14.57 g.</em>
D) The eyes in the tree are reflecting light from the torch into the man's eyes.
A mol=6.02*10^23
so
distance from earth to moon is 230,100 miles
1 big mac is about 2 inches
1 mile=5280 feet=63360inches
230,100 times 63360=14,579,136,000 inches is distance
6.02 times 10^23 times 2=x times 14,579,136,000
divide both sides by 2
6.02 times 10^23=x times 7,289,568,000
divide both sides by 6.02
10^23=x times 1210891694.35159
divide both sides by 1210891694.35159
825583769024485.8
about 825,583,769,024,486 times
Answer:
beryllium has a higher ionization energy because its radius is smaller. boron has a higher ionization energy because its radius is smaller.
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
