The formula of compound is LiClO4.3H2O
<em><u>calculation</u></em>
- <em><u> </u></em>find the mole of each element
that is moles for Li,Cl,O and that of H2O
- moles = % composition/ molar mass
For Li = 4.330/ 6.94 g/mol= 0.624 moles
Cl=22.10/35.5=0.623 moles
39.89/16 g/mol =2.493 moles
H20= 33.69/18 g/mol= 1.872 moles
- find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)
that is for Li= 0.624/0.623= 1
Cl= 0.623/0.623=1
O = 2.493/0.623 =4
H2O= 1.872/0.623=3
<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />
Answer:
I got you... 2 amino acids are linked to each other by a peptide linkage. A peptide linkage is formed when carboxyl group of one amino acid combine with the amine group of the other and during this process, a water molecule is removed.
The given tripeptide will have 2 peptide bonds. To draw the structure of given tripeptide, we will arrange them in the given order and then we will remove 2 water molecules to form 2 peptide bonds.
Explanation:
Hope this helps:)
To find this, we will use this formula:
Molar mass of element
------------------------------------ x 100
Molar mass of compound
So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.
Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol
When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!
95.994
-----------
164.086
When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.
So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
<h3>
<u>moles of H2SO4</u></h3>
Avogadro's number (6.022 × 1023) is defined as the number of atoms, molecules, or "units of anything" that are in a mole of that thing. So to find the number of moles in 3.4 x 1023 molecules of H2SO4, divide by 6.022 × 1023 molecules/mole and you get 0.5646 moles but there are only 2 sig figs in the given so we need to round to 2 sig figs. There are 0.56 moles in 3.4 x 1023 molecules of H2SO4
Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:
(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:
<h3 />
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
