That will make a gold-202 nucleus.
<h3>Explanation</h3>
Refer to a periodic table. The atomic number of mercury Hg is 80.
Step One: Bombard the with a neutron . The neutron will add 1 to the mass number 202 of . However, the atomic number will stay the same.
- New mass number: 202 + 1 = 203.
- Atomic number is still 80.
.
Double check the equation:
- Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.
Step Two: The nucleus loses a proton . Both the mass number 203 and the atomic number will decrease by 1.
- New mass number: 203 - 1 = 202.
- New atomic number: 80 - 1 = 79.
Refer to a periodic table. What's the element with atomic number 79? Gold Au.
.
Double check the equation:
- Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.
A gold-202 nucleus is formed.
The amount of CD is 35 g, and the amount of AC is 40 g.
Answer:
Partial Pressure Of Water Vapor Is 18.65mmHg At 21C0. ... How many liters of H2 gas , collected over water at an atmospheric pressure of 752 mmHg and a temperture of 21 Co, can made from 3.566 g of Zn
Explanation:
Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka =
Replacing:
1,8x10⁻⁵ =
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!
7. An exothermic reaction
8. The bonds are forming