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 quick deployment of the bag which I think is quite similar to a cushion and a balloon. It holds your head/face away from any hard surfaces. Although it does not protect your legs or other extremities that are out of the bags range.
I’d say for the answer 13.13 mmHg?
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
From the conversion of units:
1 cm^3 is equivalent to 1 mL
1 L is equivalent to 1000 mL
therefore,
to convert from liter to cm^3, we simply multiply by 1000.
Note that the multiplication will be done in the denominator.
Based on this:
density = (0.625 g/l) x (1g/1000 cm^3) = <span> 0.000625 g/cm^3
= 6.25 x 10^-4 g/cm^3</span>
