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:
2) 0.4 mol
Explanation:
Step 1: Given data
- Volume of the solution (V): 500 mL
- Molar concentration of the solution (M): 0.8 M = 0.8 mol/L
Step 2: Convert "V" to L
We will use the conversion factor 1 L = 1000 mL.
500 mL × 1 L/1000 mL = 0.500 L
Step 3: Calculate the moles of KBr (solute)
The molarity is the quotient between the moles of solute (n) and the liters of solution.
M = n/V
n = M × V
n = 0.8 mol/L × 0.500 L = 0.4 mol
 
        
             
        
        
        
Answer:
eg=linear, mg=linear
Explanation:
First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.
CO2 contains four regions of electron density and two bonding groups. For a specie containing two bonding groups, a linear molecular geometry is expected with an angle of 180°.
For a specie having two bonding groups and no lone pairs with multiple bonds, the expected electron geometry is also linear.
 
        
             
        
        
        
Answer:
-608KJ/mol
Explanation:
3 C2H2(g) -> C6H6(g)
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= ΔHC6H6 - 3ΔHC2H2
ΔHrxn = 83 - 3(230)
ΔHrxn = -608