Answer:
D. exosphere is the outer layer of the thermosphere
C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g
Answer:
1.12M
Explanation:
Given parameters:
Volume of solution = 2.5L
Mass of Calcium phosphate = 600g
Unknown:
Concentration = ?
Solution:
Concentration is the number of moles of solute in a particular solution.
Now, we find the number of moles of the calcium phosphate from the given mass;
Formula of calcium phosphate = Ca₃PO₄
molar mass = 3(40) + 31 + 4(16) = 215g/mol
Number of moles of Ca₃PO₄ =
= 2.79moles
Now;
Concentration =
Concentration =
= 1.12M
Given :
Number of molecules of hydrogen peroxide, N = 4.5 × 10²².
To Find :
The mass of given molecules of hydrogen peroxide.
Solution :
We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.
So, number of moles of hydrogen peroxide is :

Now, mass of hydrogen peroxide is given as :
m = n × M.M
m = 0.0747 × 34 grams
m = 2.54 grams
Hence, this is the required solution.