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3 years ago

A large Ka favors the _____.

Chemistry

2 answers:

Darya [45]3 years ago

8 0

The large Ka favors the <span>production of hydronium ions</span>

SVETLANKA909090 [29]3 years ago

6 0

The first one, "Production of hydronium ions."

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Repulsion is a force or interaction among the atoms in a molecule in which the atoms attract one another.

Anni [7]

Repulsion is a force or interaction among the atoms in a molecule in which the atoms REPEL one another. By the word itself repulsion, meaning to repel, it automatically bans the idea of ATTRACTION. Therefore, the statement above is FALSE. The answer to this question is "B. False".

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3 years ago

Calculate the density of a 20.015 g object that occupies 5.44 cm3 .

Scilla [17]

Answer:

$density = \frac{mass}{volume} \\ \\ density = \frac{20.015}{5.44} \\ \\ { \boxed{ \boxed{density = 3.679 \: g {cm}^{ - 3} }}}$

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3 years ago

Read 2 more answers

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

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3 years ago

Carbon dioxide :increased or decreased <br> why and when

PIT_PIT [208]

During the day or in spring and summer, plants take up more carbon dioxide through photosynthesis than they release through respiration , and so concentrations of carbon dioxide in the air decrease. Then at night or during autumn and winter, plants reduce or even stop photosynthesising, releasing carbon dioxide back into the air. This is often called the natural carbon cycle.

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2 years ago

One cubic millimeter (mm3) of blood contains 7.0 x 106 red blood cells. How many red blood cells are in 1.0 L of blood?

just olya [345]

The question gives us a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. so, to find how many red blood cells we have, we first need to convert the Liters to cm3 using this conversion factor--> 1 mL= 1 cm3

i have attached how to solve this problem.

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3 years ago