Answer:
Tile 52 and three are incorrect
Explanation:
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
B. To Identify the half reactions for the equation
Answer: Option (A) is the correct answer.
Explanation:
Force acting on a dam is as follows.
F = 
.......... (1)
Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.
F' = 
F' = 
........... (2)
Now, dividing equation (1) by equation (2) as follows.
= 
Cancelling the common terms we get the following.
= 
4F = F'
Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
