Answer: the new force will be 1/9 of the original force
Explanation: Please see attachment for explanation
First find the mass of <span>solute:
Molar mass KNO</span>₃ = <span>101.1032 g/mol
mass = Molarity * molar mass * volume
mass = 0.800 * 101.1032 * 2.5
mass = 202.2064 g of KNO</span>₃
<span>To prepare 2.5 L (0800 M) of KNO3 solution, must weigh 202.2064 g of salt, dissolve in a Beker, transfer with the help of a funnel of transfer to a volumetric flask, complete with water up to the mark, capping the balloon and finally shake the solution to mix.</span>
hope this helps!
Answer:
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
Explanation:
Let's consider the reaction between acetic acid and strontium hydroxide. This is a neutralization reaction, in which an acid reacts with a base to form salt and water. The unbalanced equation is:
HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
We have 1 acetate ion to the left and 2 to the right, so we will multiply HC₂H₃O₂(aq) by 2.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + H₂O
Finally, we multiply water by 2 to get the balanced equation.
2 HC₂H₃O₂(aq) + Sr(OH)₂(aq) ⇒ Sr(C₂H₃O₂)₂(aq) + 2 H₂O
Answer:
D
Gratitude allows for a person to gain weight.
Answer:
New pH = 3.84
Explanation:
First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.
Let's determine the moles of each compound:
0.23 M . 1.3L = 0.299 moles of NaBz
0.38 M . 1.3L = 0.494 moles of HBz
We add 0.058 of HCl, which is the same as 0.058 moles of H⁻
HCl → H⁺ + Cl⁻
As we add the moles of protons, these are going to react to the Bz⁻
In the buffer system we have these dissociations:
NaBz → Na⁺ + Bz⁻
HBz → H⁺ + Bz⁻
So, as we add protons, we have a new equilibrium:
Bz⁻ + H⁺ ⇄ HBz
In 0.299 0.058 0.494
Eq 0.241 - 0.552
Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation
pH = pKa + log (Bz⁻/HBz)
pH = 4.20 + log (0.241 / 0.552) → 3.84
