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2 years ago

Why does blowing carbon dioxide gas into aqueous barium hydroxide reduce?

2 answers:

8 0

Barium hydroxide is soluble, which signifies that it will produce ions in the solution. Supplementing carbon dioxide will result in a precipitation reaction with Ba(OH)2, which is as follows:

Ba(OH)₂ (aq) + CO₂ (g) → BaCO₃ (s) + H₂O (l)

As the ions are withdrawn from the solution, there is a reduction in the conductivity, due to this blowing carbon dioxide gas into aqueous barium hydroxide reduce.

Oksana_A [137]2 years ago

5 0

<span>If one chooses to blow carbon dioxide gas into aqueous barium hydroxide it will reduce. This happens because the ions become removed from the solution and in turn this decreases the conductivity of it.</span>

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A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production:

fgiga [73]

Answer:

d. Heat is released from the reaction

Explanation:

A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.

5 0

2 years ago

How many grams of CaF2 are present in 1.25 L of a 0.15 M solution of CaF2? How do I find the grams I am confused on that part?

dybincka [34]

Answer:

Mass = 14.64 g

Explanation:

Given data:

Volume of solution = 1.25 L

Molarity of Solution = 0.15 M

Mass of CaF₂ = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

We will calculate the number of moles of CaF₂ and then determine the mass by using number of moles.

0.15 M = number of moles of solute / 1.25 L

number of moles of solute = 0.15 M × 1.25 L

number of moles of solute = 0.1875 mol/L × L

number of moles of solute = 0.1875 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.1875 mol ×78.07 g/mol

Mass = 14.64 g

7 0

2 years ago

Which of the following statements concerning hydrocarbons is/are correct?

Alona [7]

Answer:

1. Saturated hydrocarbons may be cyclic or acyclic molecules.

2. An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0

2 years ago

Standard Hydrogen Electrode consists of platinum wire fused in a glass tube and a platinum plate coated with finely divided plat

iris [78.8K]

Answer:

To increase surface area of the platinum electrode which results in superior quality and action of the electrodes as opposed to normal platinum electrodes.

Explanation:

Platinization of Platinum is the process of covering platinum electrode with a layer of platinum black. Platinum black is a finally divided form of platinum, optimized for catalysing the addition of hydrogen to unsaturated organic compound. This increases the surface area of the platinum electrodes and therefore exhibits action superior to that of normal electrodes.

5 0

2 years ago