Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
Answer:
20 L of Cl₂
Solution:
The reaction is as follow,
H₂C₂ + 2 Cl₂ → H₂C₂Cl₄
According to equation,
167.84 g (1 mole) H₂C₂Cl₄ is produced by = 44.8 L (2 mole) of Cl₂
So,
75 g of H₂C₂Cl₄ will be produced by = X L of Cl₂
Solving for X,
X = (44.8 L × 75 g) ÷ 167.84 g
X = 20 L of Cl₂
The toxic gar expelled from the reaction between gasoline and oxygen in the vehicle's engine is both Carbon dioxide and monoxide
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
Answer:
Explanation:according to question:
. Nacl (aq) + AgNO3 (aq) --> AgCl
(s) + NaNO3 (aq).balanced.
