Answer:- 14.9 M
Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.
Density of the solution is given as 0.90 grams per mL.
From the mass and density we could calculate the volume of the solution as:
= 111 mL
Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.
= 0.111 L
Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.
= 1.65 mole
To calculate the molarity we divide the moles of ammonia by the liters of solution:
= 14.9 M
So, the molarity of the given commercial sample of ammonia is 14.9 M.
Answer:
2.63*10^23
Explanation:
1 mol rhodium = 102.91
44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =
2.63*10^23
because it is mostly sunny there
Answer:
because of catenation of carbon.
Explanation:
Catenation is the binding of an element to its self through covalent bonds to form chain or ring molecules. carbon is able to form continuous links with other carbon atoms which is the reason for the existence of a large number of organic compounds.
Answer:
113.8g
Explanation:
Statement of problem: mass of 1.946mole of NaCl
Given parameters:
Number of moles of NaCl = 1.946mole
Unknown: mass of NaCl
Solution
To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:
Number of moles = 
To find the molar mass of NaCl:
the atomic mass of Na = 23g
atomic mass of Cl = 35.5g
Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹
Mass of NaCl = Number of moles x molar mass of NaCl
Mass of NaCl = 1.946 x 58.5 = 113.8g
