The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
28.01g
Explanation:
Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.
The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .
The molecular weight of the compound CO is therefore
equal to the sum of the weight of both elements.
That’s = 12.01g + 16.00g
= 28.01g
Therefore, the molecular weight of CO is 28.01g
Here is an image of the periodic table. Hope this helps.
Answer:
(n, l, m sub l, m sub s)
N: principle quantum number (1,2,3,4,etc)
l: angular momentum quantum number, the shape (l has to be at least 1 less than n, but can be 0 depending on n)
M sub l: magnetic quantum number (l determines this number)
M sub s: spin quantum number (can only ever be 1/2 or -1/2)
Explanation: