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MakcuM [25]
3 years ago
7

The pH of a solution is 3.81. What is the OH concentration in the solution?​

Chemistry
1 answer:
disa [49]3 years ago
3 0

Answer:

6.46 × 10⁻¹¹ M

Explanation:

Step 1: Given data

pH of the solution: 3.81

Step 2: Calculate the pOH of the solution

We will use the following expression.

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 3.81 = 10.19

Step 3: Calculate the concentration of OH⁻ ions

We will use the definition of pOH.

pOH = -log [OH⁻]

[OH⁻] = antilog -pOH = antilog -10.19 = 6.46 × 10⁻¹¹ M

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Answer: the answer is B

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A plate moves 200 m in 10,000 years. What is its rate in cm/year?
Varvara68 [4.7K]

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<h2>2 cm/year</h2>

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3 0
3 years ago
Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.
velikii [3]
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
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3 years ago
Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no rea
iragen [17]

Answer :

(A) The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

(B) The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

(C) The balanced molecular equation will be:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

(D) The balanced molecular equation will be:

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}

Explanation :

Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

Part A  : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

Part B : lithium sulfate and lead(II) acetate

The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.

Part C : copper(II) nitrate and sodium sulfide

The balanced molecular equation will be:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.

Part D : strontium nitrate and potassium iodide

The balanced molecular equation will be:

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+SrI_2(aq)

In this reaction, strontium iodide and potassium nitrate are soluble solution.

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}

6 0
3 years ago
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