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creativ13 [48]
2 years ago
8

The oh concentration in a 1.0 x10 3 m ba oh 2 solution is

Chemistry
2 answers:
Lisa [10]2 years ago
4 0
Ba(OH)2 dissociates according to the equation below to yield Barium ions and hydroxide ions.
Ba(OH)2 = Ba²⁺ + 2 OH⁻
The concentration of Ba²⁺ is 1.0 ×10^-3 M
Thus that of OH⁻ ions will be 2× 1.0 ×10^-3 = 2.0 × 10^-3 M
Thus; the answer is 2.0 × 10^-3 M
shusha [124]2 years ago
3 0

The {\text{O}}{{\text{H}}^ - } concentration in a 1.0 \times {10^{ - 3}}{\text{ M Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} solution is \boxed{2.0 \times {{10}^{ - 3}}{\text{ M}}}.

Further Explanation:

Concentration is used to describe relationship between various components of solution. For this purpose, a wide variety of concentration terms are used. Some of these are enlisted below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

7. Parts per billion (ppb)

Molarity is defined as moles of solute present in one litre of solution. It is represented by M and its unit is mol/L. The expression for molarity of solution is as follows:

{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}

{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} is a strong base and it dissociates into its respective ions as follows:

{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} \rightleftharpoons {\text{B}}{{\text{a}}^{2 + }} + 2{\text{O}}{{\text{H}}^ - }  

This indicates one mole of {\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} dissociates to produce one mole of {\text{B}}{{\text{a}}^{2 + }}  ions and two moles of {\text{O}}{{\text{H}}^ - } ions.

Given information:

 {\text{Concentration of Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} = 1.0 \times {10^{ - 3}}{\text{ M}}

Since one mole of {\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} forms two moles of {\text{O}}{{\text{H}}^ - } ions, concentration of {\text{O}}{{\text{H}}^ - } ions becomes twice the concentration of {\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} and is calculated as follows:

\begin{aligned}{\text{Concentration of O}}{{\text{H}}^ - } &= 2\left( {1.0 \times {{10}^{ - 3}}{\text{ M}}} \right) \\&= 2.0 \times {10^{ - 3}}{\text{ M}}\\\end{aligned}  

Therefore concentration of {\text{O}}{{\text{H}}^ - } ions comes out to be 2.0 \times {10^{ - 3}}{\text{ M}}.

Learn more:

  1. Calculation of volume of gas: brainly.com/question/3636135
  2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, Ba(OH)2, OH-, Ba2+, 1.0*10^-3 M, 2.0*10^-3 M, molarity of solution, moles of solute, volume of solution, mole fraction, parts per million, parts per billion, mass percent, volume percent.

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How many moles of CO2 could fit in a 475 mL bag a -22° C and 855 mmHg?
Sergeeva-Olga [200]

Answer:

The amount of moles of CO₂ is 0.026 moles

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The equation known as the ideal gas equation explains the relationship between the four variables P (Pressure), V (Volume), T (Temperature) and n (Amount of substance). The ideal gas law is expressed mathematically as:

P*V = n*R*T

where P represents the pressure of the gas, V its volume, n the number of moles of gas (which must remain constant), R the constant of the gases and T the temperature of the gas.

In this case:

  • P= 855 mmHg
  • V= 475 mL= 0.475 L
  • n= ?
  • R= 62.36367 \frac{mmHg*L}{mol*K}
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Replacing:

855 mmHg*0.475 L=n*62.36367 \frac{mmHg*L}{mol*K} *251 °K

Solving:

n=\frac{855 mmHg*0.475 L}{62.36367 \frac{mmHg*L}{mol*K} *251 K}

n= 0.026 moles

<u><em>The amount of moles of CO₂ is 0.026 moles</em></u>

4 0
2 years ago
If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces
natta225 [31]
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
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7 0
2 years ago
1. What volume of toluene do you need to add to 1 mL of ethyl acetate to make an equi-molar mixture? The density of ethyl acetat
galben [10]

Answer:

1.06 mL of toluene will be needed.

Explanation:

Equi-molar mixture means equal moles of all the components.

as given the volume of ethyl acetate = 1mL

Density of ethyl acetate = 0.898 g/mL

The relation between density, mass and volume is :

density=\frac{mass}{volume}

mass=volumeXdensity

mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams

the moles are related to mass as:

moles=\frac{mass}{molarmass}

For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol

moles of ethyl acetate will be:

moles=\frac{0.898}{88}= 0.01mol

So we need 0.01 moles of toluene also

For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene

mass required = 0.01 X 92=0.92grams

for 0.92 grams of toluene volume required will be:

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4 0
2 years ago
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oksian1 [2.3K]

Answer:

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During the night, and specially in the early morning, before dawn, the temperature of the air descends, and part of the vapor in the air condensates in tiny droplets that accumulate over the surface of the plant's leaves, and other solid surfaces like the winshields and hoods of the cars.

Then, the phase transition that occurs is from gas (vapor) to liquid, which is called condensation and represented with the arrow labeled 4 on the diagram.

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tankabanditka [31]
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2 years ago
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