Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Metallic bonding
Metals consist of giant structures of atoms arranged in a regular pattern. The electrons from the outer shells of the metal atoms are delocalised , and are free to move through the whole structure. This sharing of delocalised electrons results in strong metallic bonding .
Answer:
a) nitrogen
b) nitrogen =5
Oxygen = 6
Fluorine =7
Explanation:
Usually, if we have two or more elements in a compound, the central atom in the compound is the atom having the least value of electro negativity.
If we consider fluorine, oxygen and nitrogen; nitrogen is the least electronegative of the trio hence it should be the central atom of the triatomic molecule.
The number of valence electrons on the valence shell of each atom is shown below;
nitrogen =5
Oxygen = 6
Fluorine =7
Explanation:
the molar mass of the compound is 1502g/mol
