Atomic mass Ca = 40.078 a.m.u
40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??
72.8 x ( 6.02x10²³) / 40.078 =
4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms
hope this helps!
POH value was calculated by the negative logarithm of hydroxide ion concentration.
To know the hydrogen ion concentration, we need to know the pH value, that can be found out if pOH is known
pH + pOH = 14
pH = 14 - pOH
pH = 10.65
once the pH is known we have to find the antilog.
[H⁺] = antilog (-pH)
antilog can be found by
[H⁺] = 10^(-10.65)
[H⁺] = 2.2 x 10⁻¹¹ M
Answer:
0.500 mole of Xe (g) occupies 11.2 L at STP.
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Mole ratio
- Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
0.500 mole Xe (g)
<u>Step 2: Convert</u>
- [DA] Set up:
- [DA] Evaluate:
Topic: AP Chemistry
Unit: Stoichiometry
Potassium is an alkali metal with the chemical symbol K. It has an atomic number of 19, meaning that it has 19 positively charged protons. It also contains 19 electrons, which have a negative charge, and 20 neutrons, which do not hold a charge
Hope this help
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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