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UkoKoshka [18]
3 years ago
10

A solution contains 1 LaTeX: \times\:×10−4 M OH– ions. Calculate the solution pH value, and determine if the solution is acidic,

basic, or neutral. Helpful math formulas: LaTeX: pH=-\log\left[H_3O^+\right]p H = − log ⁡ [ H 3 O + ] LaTeX: pOH=-\log\left[OH^-\right]p O H = − log ⁡ [ O H − ] LaTeX: pH+pOH=14.00p H + p O H = 14.00 LaTeX: \left[H_3O^+\right]\left[OH^-\right]=1.0\times10^{-14}
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
5 0

<u>Answer:</u> The pH value of the solution is 10 and the solution is basic in nature.

<u>Explanation:</u>

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=1\times 10^{-4}M

Putting values in above equation, we get:

pOH=-\log(1\times 10^{-4})\\\\pOH=4

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-4=10

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

  • The pH range of acidic solution is 0 to 6.9
  • The pH range of basic solution is 7.1 to 14
  • The pH of neutral solution is 7.

As, the pH of the solution is 10 and is lying in the range of basic solution, so the solution is basic in nature.

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How are Parkinson's disease and ALS similar?
Diano4ka-milaya [45]

Answer:

Both diseases affect the control of voluntary muscles.

Explanation:

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ALS (amyotrophic lateral sclerosis) is a progressive nervous system disease that affects nerve cells in the brain and spinal cord. The first symptoms usually involve muscle weakness, and as the disease progresses, it results in the loss of muscle control.

Scientists don't know the exact cause of these diseases. As the cause is not known, there is no exact way to prevent them. There is no cure for them, either. The treatment is focused on the management of symptoms.

This is why the third option is the correct one.

4 0
3 years ago
A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
balu736 [363]
PH + pOH = 14

11.8 + pOH = 14

pOH = 14 - 11.8

pOH = 2.2

[OH-] = 10 ^- pOH

[OH-] = 10 ^- 2.2

[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B

hope this helps!





3 0
3 years ago
Read 2 more answers
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
A certain system absorbs 350 joules of heat and has 230 joules of work done on it what is the value of
marysya [2.9K]
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate). 

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 

(10.0 g) / (0.0480 mol) = 208.3 g/mol 

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
7 0
3 years ago
N2-3H2 → 2NH3
aniked [119]

Answer:

6 mols H2 are needed

Explanation:

N2 = 28.01g/mol

H2 = 2.02g/mol

\frac{2 mol N_{2} }{1} * \frac{3 mol H_{2}  }{ 1 mol N_{2} } = 6 mol H2

8 0
3 years ago
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