The question is typed wrong. Assuming the correct question is
How many liters of 0.98 M H2SO4 solution would react completely with 3.5 moles Ca(OH)2 ?
Answer:-
3.571 litres
Explanation:-
The balanced chemical equation for this reaction is
H2SO4 + Ca(OH)2 = CaSO4 + 2 H2O
From the balanced chemical equation we see that
1 mole of Ca(OH)2 reacts with 1 mol of H2SO4.
∴3.5 moles of Ca(OH)2 reacts with 1 x 3.5 / 1 = 3.5 mol of H2SO4.
Strength of H2SO4 = 0.98 M
Volume of H2SO4 required = Number of moles of H2SO4 / Strength of H2SO4
= 3.5 moles / 0.98 M
= 3.571 litre
Hydrogen bonding occurs when hydrogen is bonded to an oxygen or nitrogen or fluorine atom. In this case, the hydrogen atom in a hydrogen fluoride molecule will be able to bond to the fluoride atom of another hydrogen fluoride molecule, forming a hydrogen bond.
Answer:
Water has a pH of 7 - its not acid or base (its neutral {in the middle})
Explanation:
Since the pH scale maximum number is 14, the middle number - is 7 - which is where water is
On the off chance that the red blood cells are smaller than ordinary, this is called microcytic anemia. The significant reasons for this sort are low-level iron, anemia, thalassemia.
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂
Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction
Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
