Answer:
Percent Composition of Compounds
The percent composition (by mass) of a compound can be calculated by dividing the mass of each element by the total mass of the compound.
LEARNING OBJECTIVES
Translate between a molecular formula of a compound and its percent composition by mass
- The atomic composition of chemical compounds can be described in a variety of ways, including molecular formulas and percent composition.
- The percent composition of a compound is calculated with the molecular formula: divide the mass of each element found in one mole of the compound by the total molar mass of the compound.
- The percent composition of a compound can be measured experimentally, and these values can be used to determine the empirical formula of a compound.
- percent by mass: The fraction, by weight, of one element of a compound.
- The atomic composition of chemical compounds can be described using a variety of notations including molecular, empirical, and structural formulas. Another convenient way to describe atomic composition is to examine the percent composition of a compound by mass.
- Percent Composition by Mass
Percent composition is calculated from a molecular formula by dividing the mass of a single element in one mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.
Explanation:
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Answer:
Explanation:
Hello!
In this case, since the molarity of a solution is defined as the moles of solute divided by the volume of solution liters, given the mass of potassium sulfate, we can compute the moles by using its molar mass (174.24 g/mol):
Thus, since one mole of potassium sulfate has two moles of potassium ions (K₂) and one mole of sulfate ions, we can compute the moles of each ion as shown below:
In such a way, the molarity of each ion turns out:
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Answer:
The answer is 12.35
Explanation:
From the question we are given that the concentration of 
is 
Generally The rate equation is given as
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
and 
the rate constant has a value 
Substituting and making [
] the subject we have
![[OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%20%3D%20%5Cfrac%7B1%20%2A%2010%5E%7B-14%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B1%20%2A%2010%5E%7B-14%7D%7D%7B8.1%20%2A10%5E%7B-6%7D%7D%20%3D1.235%20%2A%2010%5E%7B-9%7D)
![[OH ^ {-}] = 1.235 * 10^{-9}M](https://tex.z-dn.net/?f=%5BOH%20%5E%20%7B-%7D%5D%20%3D%201.235%20%2A%2010%5E%7B-9%7DM)
Multiply the value by 
as instructed from the question we have
Answer = 
Hence the answer in 2 decimal places is 12.35
Answer:
185 gms o NaBr
Explanation:
Na Br mole weight = 22.989 +79.904 = <u>102.893 gm/mole</u>
1.083 x 10^24 molecules / 6.022 x 10^23 molecules / mole = <u>1.798 moles</u>
102.893 * 1.798 = 185 gms
