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:

Explanation:
Hello there!
In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

And can be solved for the total pressure as follows:

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

Then, we can plug in to obtain the total pressure:

Regards!
Answer: 2.60 x 10^23 molecules
50.0 grams x (1 mol/115.79 grams) = 0.431816219 moles
0.431816219 mol x (6.02 x 10^23 molecules/1 mol) = 2.599533638 x 10^23 molecules (final answer is rounded)
The following is the introduction to a special e-publication called Determining the Age of the Earth (click the link to see a table of contents). Published earlier this year, the collection draws articles from the archives of Scientific American. In the collection, this introduction appears with the title, “Stumbling Toward an Understanding of Geologic Timescales.”
1) To find the change in enthalpy, determine the difference between the potential energy of the products and the potential energy of the reactants. (on this diagram, C-A) To find the activation energy, find the difference between the potential energy of the reactants and the "peak" of the curve (on this diagram, B-A). For this diagram, both the enthalpy and activation energy are positive.
2) If the reaction was exothermic, enthalpy would be negative, and the potential energy of the reactants would be greater than the potential energy of the products.