You did not specify the types, but I believe the answer would be gamma radiation.
Answer:
There are three significant figures
Explanation:
When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.
The empirical formula : C₂Cl₇
The molecular formula : C₁₀Cl₃₅
<h3>Further explanation</h3>
Given
8.81 g Carbon
91.2 g Chlorine
Molar Mass: 1362.5 g/mol
Required
The empirical formula and molecular formula
Solution
Mol ratio :
C = 8.81 g : 12.011 g/mol =0.733
Cl = 91.2 g : 35,453 g/mol = 2..572
Divide by 0.733
C : Cl = 1 : 3.5 = 2 : 7
The empirical formula : C₂Cl₇
(The empirical formula)n = the molecular formula
(C₂Cl₇)n = 1362.5
(2x12.011+7x35.453)n=1362.5
(272.193)n=1362.5
n = 5
Answer:
Ppm <u>means the concentration of vapors or gases</u> expressed in parts per million of polluted air.
Explanation:
In Chemistry and Physics, part per million or PPM is a measure that expresses the number of units of a certain existing substance for every million units of the mixture. (Concentration)
In climate the equation is used so:
- ppm = parts of polluted air / million parts of air
.
- ppm = Liters of polluted air / 10∧6 Liters of air.
This is used to find the concentration of gases expressed as parts of gas per million parts of the
polluted air.
In this case, it would be<u> 407.4L of CO2 per 10∧6L of air.</u>
The empirical formula of the compound is calculated as follows
first calculate the mass of oxygen= 12-(4.09 +3.71)= 5.02g
then calculate the moles of each element, moles = mass/ molar mass
moles of K = 4.09g/39 g/mol(molar mass of K) = 0.105 moles
moles of Cl = 3.71g/35.5 g/mol(molar mass of Cl) = 0.105 moles
moles of O = 5.02g/ 16g/mol(molar mass of O) = 0.314 moles
then calculate e mole ratio by dividing each mole by the smallest number of moles ( 0.105 moles)
K=0.105/0.105= 1
Cl=0.105 /0.105=1
O= 0.314/0.105=3
therefore the empirical formula = KClO3
