Answer:
length
Explanation:
cm measures length. Think of a ruler.
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:
Mantle
Explanation:
Convection currents in the Earth occur in the mantle. The core of the Earth is extremely hot, and material in the mantle close to the core is heated...
Hope that help <3
<h3>
Answer:</h3>
2.809 L of H₂SO₄
<h3>
Explanation:</h3>
Concept tested: Moles and Molarity
In this case we are give;
Mass of solid sodium hydroxide as 13.20 g
Molarity of H₂SO₄ as 0.235 M
We are required to determine the volume of H₂SO₄ required
<h3>First: We need to write the balanced equation for the reaction.</h3>
- The reaction between NaOH and H₂SO₄ is a neutralization reaction.
- The balanced equation for the reaction is;
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
<h3>Second: We calculate the umber of moles of NaOH used </h3>
- Number of moles = Mass ÷ Molar mass
- Molar mass of NaOH is 40.0 g/mol
Moles of NaOH = 13.20 g ÷ 40.0 g/mol
= 0.33 moles
<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
- From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
- Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
- Thus, Moles of H₂SO₄ = moles of NaOH × 2
= 0.33 moles × 2
= 0.66 moles of H₂SO₄
<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
- When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
- That is; Volume = Number of moles ÷ Molarity
In this case;
Volume of the acid = 0.66 moles ÷ 0.235 M
= 2.809 L
Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.
