Answer:
About 0.0940 M.
Explanation:
Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:
Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:
Therefore, the molarity of the original solution was:
![\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%20%5Ctext%7BHA%7D%5Cright%5D%20%3D%20%5Cfrac%7B0.00188%5Ctext%7B%20mol%7D%7D%7B20.00%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%3D%200.0940%5Ctext%7B%20M%7D)
In conclusion, the molarity of the unknown acid is about 0.0940 M.
The period of the life cycle during which positive nitrogen balance is most likely to occur is childhood.
<h3>What is nitrogen balance?</h3>
The term nitrogen balance refers to the fact that there is a balance between the intake and the loss of nitrogen. The intake of nitrogen occurs when a person takes in food that contain proteins which are a rich source of nitrogen in the body and helps in the process of the build up of the cells in the body.
The period of the life cycle during which positive nitrogen balance is most likely to occur is childhood.
Learn more about nitrogen balance:brainly.com/question/14570903
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Answer:
B
Explanation:
produce a product improve it identify a problem make a solution
81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side.
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
