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Dafna11 [192]
2 years ago
12

Joanna is performing a reaction that generates a moderate amount of hydroxide. To mimic biological conditions (most bodily fluid

s in living organisms retain a relatively constant pH around 7), she knows that she needs to add a buffer to the reaction. Which of the following will MOST effectively neutralize the OH− ions produced by the reaction
Chemistry
1 answer:
Nady [450]2 years ago
7 0

Answer:

As for your question, I know to forget to put the options, specifically that your question is incomplete.

Explanation:

Although it could help you by telling you that always a reaction that seeks to balance the pH, and achieve neutrality ... It is necessary to achieve a concentration of OH equal to that of H +, in this way the hydroxyl and the protons.

You might be interested in
in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?
djverab [1.8K]
In collecting the precipitate, it is inappropriate to heat <span>the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.</span>
6 0
2 years ago
Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0
2 years ago
What is the value of Keq for the reaction expressed in scientific notation? 2.1 x 10-2 2.1 x 102 1.2 x 103 1.2 x 10-3
Lady_Fox [76]

Complete question:

Consider the reaction.

At equilibrium at 600 K, the concentrations are as follows.

2HF -----> H₂ + F₂

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3 M

[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?

2.1 x 10-2

2.1 x 102

1.2 x 103

1.2 x 10-3

Answer:

2.1 × 10^-2

Explanation:

Kequilibrum(Keq) = product/reactant

Equation for the reaction :

2HF -----> H₂ + F₂

Therefore,

Keq = [H2][F2] / [HF]^2

Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2

Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]

Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]

Keq = 2.0665 × 10^(-6 - (-4))

Keq = 2.0665 × 10^(-6 + 4)

Keq = 2.1 × 10^-2

7 0
2 years ago
Read 2 more answers
If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln
Ksju [112]
Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
5 0
2 years ago
Read 2 more answers
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
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