Answer:

Explanation:
The molar mass is the mass of a substance in grams per mole.
To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.
The compound given is:

The compound has 1 Ca (calcium) and 2 Cl (chlorine).
Mass of Calcium
- The molar mass of calcium is 40.08 g/mol
- There is only one atom of Calcium in CaCl₂, so the number above is what we will use.
Mass of Chlorine
- The molar mass of chlorine is 35.45 g/mol
- There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
- 35.45 * 2= 70.9 g/mol
Molar Mass of CaCl₂
- Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
- 40.08 g/mol + 70.9 g/mol =110.98 g/mol
The molar mass of CaCl₂ is <u>110.98 grams per mole. </u>
Answer: water could be used to wash it since the reaction has ended.
Explanation:
There will be no reaction of water with the Grignard reagent since the reaction has ended, as it is well known that water is a universal solvent for washing of glasswares after experiments but if it is during the reaction it will be more advisable to rinse with alcohol to enhance more accuracy during the experiment
Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
Answer:
Its mass is about the same as that of a proton
Explanation: