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
Answer: the atomic number 7 is
Nitrogen
Explanation:
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
Explanation:
It is given that r = 0.283 nm. As 1 nm = 
.
Hence, 0.283 nm = 
- Formula for coulombic energy is as follows.
where, e = 
C
= 
= 
- As 1 eV =
So, 1 J = 
Hence, U = 
= 8.9 eV
- Also, 1 J =
= 
kJ/mol
Therefore, U = 
kJ/mol
= 
