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.
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
Answer:
I think It is an organism
The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.576 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.019862 mol
(101,183.9 Pa)V = (0.019862 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L