.500 mol ( 6.02x10^23 molecules/1 mol) = 3.01x10^23
The answer to this question is:
An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (Na) and 45.24 percent fluorine (F). Determine the empirical formula of the compound?.
<span>"NaF"
</span>
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Hgibore
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Answer:
Explanation:
You need to know one piece of information for these problems. That is as follows:
There are 6.02 x 10^23 atoms in a mole of atoms or 6.02 x 10^23 molecules in a mole of molecules (actually there are 6.02 x 10^23 in a mole of anything).
So in 3.161 x 10^21 molecules of CO2 there are 3.161 molecules x (1 mole CO2/6.02 x 10^23 molecules CO2)= ?? moles CO2. Then multiply that by 2 to find the moles of O in CO2.
The others are done th same way.
Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
Answer:If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2.
If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.
Explanation:
