Answer:
Explanation:every feet has 12in so divde, 67/12 so its 5ft 7/12
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
More about buffer: brainly.com/question/4177791
#SPJ4
Answer:
3.5 × 10⁵ g of salt
Explanation:
<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>
We have this data:
- 1.000 mol salt is equal to 58.44 g salt
- 1.0 L of ocean water contains 0.60 mol of salt
We will need the following relations:
We can use proportions:

Answer:
The answer is
<h2>2.00 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula

From the question
actual measurement = 46.37 g
error = 47.25 - 46.37 = 0.88
The percentage error of the measurement is

We have the final answer as
<h3>2.00 %</h3>
Hope this helps you