Answer:
Br2+ 2H2O + SO2= 2HBr + H2SO4
Let's Compare the left side of the equation to the right side of the equation.
Left: Br= 2, H= 2, S= 1, O = 1+2
Right: Br=1, H= 1+2, S=1, O= 4
We can see that only S is balanced and not the other 3 elements.
I'll try to make each element balance.
For Br; I'll multiply by 2 on the left to make it equal to the right.
For H; Since the 2 for Br on the right affected also H, that H ( for HBr) Already has a 2, but then it adds with the other H2( for H2SO4) to give a total of 4 H on the right side. But then there's only 2 H on the left. so we multiply that 2 by a 2 ( which is written infront of the H2O to give a total of 4 H on the left side.
For O; Because of the 2 infront of the H2O, it affects the O in H2O..so now we have 2 O plus the 2 O ( in SO2) to give a total of 4 O which is equal to the right side.
You can wait until he either breaks up (if he does) or if just forget about him and go for another guy.
Firstly need to determine the empirical formula of the hydrocarbon. Empirical formula is the simplest whole number ratio of components of the compound. Molecular formula is the actual composition of the components in the compound.
percentage of C - 82.66%
percentage of H - (100-82.66) = 17.34 %
in 100 g of compound ;
mass of C - 82.66 g
mass of H - 17.34 g
C H
mass in 100 g 82.66 g 17.34 g
molar mass 12 g/mol 1 g/mol
number of moles 6.88 mol 17.34 mol
(mass/molar mass)
divide the number of moles by least number of moles (6.88 mol)
6.88 mol/6.88 17.34/6.88
1 2.52
multiply these by 2 to get a whole number
C - 1x 2 = 2
H - 2.52 x 2 = 5.04
round off to nearest whole number
C - 2
H - 5
ratio of C to H is 2:5
empirical formula - C₂H₅
empirical formula mass = 12 g/mol x 2 + 5 * 1 g/mol = 29 g
next have to find how many empirical units are there in the molecular unit
molecular unit mass = 58.12 g
empirical unit = 29 g
then number of empirical units = 58.12 / 29 = 2
rounded off , number of empirical units = 2
(C₂H₅) * 2 units
molecular formula = C₄H₁₀
Explanation:
Answer:
The amount of precipitate formed would 7.175 grams of silver chloride.
Explanation:
Moles of NaCl = n
Volume of NaCl solution = 50.0 mL = 0.050 L
Molarity of the hydrogen peroxide = 2.0 M
Moles of silver nitarte = n'
Volume of silver nitrate solution = 50.0 mL = 0.050 L
Molarity of the silver nitrate = 1.0 M
According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :
of NaCl
This means that silver nitrate is in limiting amount and NaCl is in excessive amount.
So, the amount of AgCl depends upon amount of silver nitrate.
According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.
Then 0.050 moles of silver nitrate will give;
of AgCl
Mass of 0.050 moles of AgCl ;
The amount of precipitate formed would 7.175 grams of silver chloride.
