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Molodets [167]
3 years ago
12

Order these species by increasing concentration of H30+ in a 1.0 M aqueous solution. (From the

Chemistry
1 answer:
masha68 [24]3 years ago
4 0

Answer:

OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃

Explanation:

We can do some rough calculations to find the approximate pH values of these solutions.

H₂CO₃

Kₐ ≈ 10⁻⁶

\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-6} \times 10^{-1}} = \sqrt{10^{-7}} = 10^{-3.5}\\\text{pH} = -\log (10^{-3.5}) = \mathbf{3.5}

NH₄⁺

Kb of NH₃ ≈ 10⁻⁵

Kₐ of NH₄⁺ ≈ 10⁻⁹

\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-9} \times 10^{-1}} = \sqrt{10^{-10}} = 10^{-5}\\\text{pH} = -\log (10^{-5}) = \mathbf{5}

OH⁻

Strong base

[OH⁻] = 10⁻¹

pOH = 1

pH = 14 - 1 = 13

HCO₃⁻

Salt of dibasic acid

K₁ ≈ 10⁻⁶; K₂ ≈ 10⁻¹⁰

{\text{H}^{+}} = \sqrt{K_{1}K_{2}} = \sqrt{10^{-6}\times 10^{-10}} = \sqrt{10^{-16}} = 10^{-8}\\\text{pH} = -\log (10^{-8}) = \mathbf{8}

NH₃

Kb ≈ 10⁻⁵

\text{OH}^{-} = \sqrt{K_{\text{b}}c} = \sqrt{10^{-5} \times 10^{-1}} = \sqrt{10^{-6}} = 10^{-3}\\\text{pOH} = -\log (10^{-3}) = 3

pOH = 14 - 3 = 11

H₂O

Neutral. pH = 7

Order from lowest [H₃O⁺] to highest [H₃O⁺]:

      OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃  

pH   1 3        11          8            7         5           3.5

 

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castortr0y [4]

Answer:

n_{Cl^-}=25molCl^-

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

5\frac{molMgCl_2}{L}

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-

Best regards.

4 0
2 years ago
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts
Y_Kistochka [10]

Answer:

Explanation:

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?

Mg + O2 → MgO (unbalanced)

first, balance the equation

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there is a 1:1 ratio of magnesium to oxygen atoms

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the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.

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kozerog [31]

Answer:

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At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

7 0
3 years ago
1.26 * 10^4 + 2.50 * 10^4 in sceintific notation
eduard
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