Formal charge can be calculated from the following formula
Formal charge = valency of central atom - (number of lone pair of electrons + number of covalent bonds)
a) for methylene:
Formal charge = 4 -( 2+ 2) = 0
b) For methyl free radical
Formal charge = 4- (3 +1) = 0
Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
When an iron is dipped in Copper Sulphate
Solution this reaction between them and
copper sulphate change into blue color to light
green color. This show that iron is more
reactive then copper, it can to replace copper
from CuSO4 , CuSO4 is of blue color and
FeSO4 is light green color.
Hope it helps
Answer:
if an atom gains an electron, the ion has negetive charge
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
