Answer:
1.63ₓ10⁻⁶ g of U
139.03 g of H
0.385 g of O
141.8 g of Pb
Explanation:
In first place, we need to convert the number of atoms to moles, as we know that 1 mol of anything occupies 6.02×10²³ particles
Therefore:
4.12×10¹⁵ atoms of U . 1 mol / 6.02×10²³ atoms = 6.84×10⁻⁹ moles of U
8.37×10²⁵ atoms of H . 1 mol /6.02×10²³ atoms = 139.03 moles of H
1.45×10²² atoms of O . 1 mol /6.02×10²³ atoms = 0.0241 moles of O
4.12×10²³ atoms of Pb . 1 mol /6.02×10²³ atoms = 0.684 moles of Pb
Moles . Molar mass = Mass (g)
6.84×10⁻⁹ moles of U . 238.03 g/mol = 1.63ₓ10⁻⁶ g of U
139.03 moles of H . 1 g/mol = 139.03 g of H
0.0241 moles of O . 16 g/mol = 0.385 g of O
0.684 moles of Pb . 207.2 g/mol = 141.8 g of Pb
The isotope U-235 is important because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and we use the expression 'nuclear fission'. Meanwhile, like all radioactive isotopes, they decay.
Answer: 66
Explanation: To find the mass number you would have to add the number of protons and the number of neutrons.
Yes. Exhale into a jar.
That's a hella ratchet way to capture CO2, but it works nonetheless
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm