The percentage mass composition of chlorine in HClO₃ is 42%.
Percentage mas of chlorine is calculated as follows:
atomic mass of chlorine = 35.5
(1) In HClO (gram-formula mass = 52 g/mol)
Percentage mass composition = 35 * 100% / 52 = 67.3%
(2) In HClO2 (gram-formula mass = 68 g/mol)
Percentage mass composition = 35 * 100%/68 = 51.4%
(3) HClO3 (gram-formula mass = 84 g/mol)
Percentage mass composition = 35 * 100% / 84 = 42%
(4) HClO4 (gram-formula mass = 100. g/mol)
Percentage mass composition = 35 * 100% / 100 = 35.5%
Therefore, the percentage mass composition of chlorine in HClO₃ is 42%.
Learn more about percentage mass composition at: brainly.com/question/18646836
None of the compounds have a percent composition of chlorine equal to 42%.
To find the compound with a percent composition of chlorine equal to 42%, we calculate the mass of chlorine in each compound and calculate the percent composition. Let's go through each option:
None of the compounds have a percent composition of chlorine equal to 42%. The correct answer is none of the above.
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Answer:
There are 1, 13 moles of chlorine gas.
Explanation:
We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:
PV= nRT ---> n= PV/RT
n= 0,98 atm x 35,5 L /0,082 l atm / K mol x 373 K
n= 1,137448506 mol
Answer:
It undergoes a physical change.
Explanation:
the frequency in Hertz
the Energy in Joules
2) NH3
3) KNO3
4) NaCl
Answer:
The correct answer is 2.
Explanation:
The NH3 solubility curve shows that when the temperature increases, the solubility of the compound decreases. The molecule is not flat but has a tetrahedral shape with a vacant vertex, and this is due to the formation of hybrid sp³ orbitals. Unlike other compounds, the NH3 solubility curve has been experimentally tested and is attached for better understanding.
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b. .78 M
c. 3.85 M
b. 4.98 M
Answer:
The correct answer is c.
Explanation:
First, look for the same problem to verify that the correct volume is 0.702 L.
Now we must calculate the amount of glucose moles (n) in the 500g:
n =
n = 2,7 mol
Molarity is given by moles of compound per liter. In this way, we calculate the molarity (M) as:
M =
M =
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