Answer:
There must be two Chlorine atoms for every one Calcium atom in order to fulfill Chlorine's octet rule and pair Calcium's unpaired electrons.
Explanation:
Calcium has two unpaired electrons in its Lewis dot structure, while Chlorine has one unpaired electron.
<em>So why can't we just make a double bond for </em><em>one</em><em> Chlorine?</em>
Chlorine has seven valence electrons, so once it shares electrons with Calcium, the octet rule is accomplished, and no more pairs can be made.
Answer:
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
Explanation:
Let's make some rules of three, to solve this problem:
1 mol of ethanol has 2 moles of C, 6 moles of H, and 1 mol of oxygen
Therefore, 0.75 moles of ethanol must have 0.75 mol of oyxgen
Let's convert the moles to mass → 0.75 mol . 16 g/ 1 mol = 12 g
1 mol of formic acid has 2 moles of H, 1 mol of C and 2 mol of oxygen
0.60 moles of formic acid must have (0.6 .2) / 1 = 1.2 mol of O
If we convert the amount to mass → 1.2 mol . 16 g/ 1mol = 19.2 g
1 mol of water has 1 mol of oyxgen
Therefore, we have 1 mol of oxygen with a mass of 16 g.
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
Answer:
0.78 atm
Explanation:
Step 1:
Data obtained from the question. This includes:
Mass of CO2 = 5.6g
Volume (V) = 4L
Temperature (T) =300K
Pressure (P) =?
Step 2:
Determination of the number of mole of CO2.
This is illustrated below:
Mass of CO2 = 5.6g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Number of mole CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 5.6/44
Number of mole of CO2 = 0.127 mole
Step 3:
Determination of the pressure in the container.
The pressure in the container can be obtained by applying the ideal gas equation as follow:
PV = nRT
The gas constant (R) = 0.082atm.L/Kmol
The number of mole (n) = 0.127 mole
P x 4 = 0.127 x 0.082 x 300
Divide both side by 4
P = (0.127 x 0.082 x 300) /4
P = 0.78 atm
Therefore, the pressure in the container is
Given that 1 micrometer or micron (um) is equivalent by definition to 1 x 10^-6 m, this means that 1 square micron (um^2) is equivalent to (1 x 10^-6)^2 m^2, or 1 x 10^-12 m^2.
(2.60 um^2) * (1 x 10^-12 m^2 / 1 um^2) = 2.60 x 10^-12 m^2
Therefore the layer of graphene covers an area of 2.60 x 10^-12 m^2.
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