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2 years ago

If the conjugate base of a molecule has a pkb of 1.4, what would you expect the molecule to be?

Chemistry

1 answer:

maw [93]2 years ago

3 0

If the conjugate base of a molecule has a pKb of 1.4, the molecule should be a Weak Acid.

Notice this question gives us the pKb of the molecule, not the pKa. Because of this, the pH scale basically gets reversed, so lower numbers in pKb correlate with stronger bases, and higher numbers in pKb correlate with stronger acids - the exact opposite of the pH scale.

It's important to make sure you completely understand the terms of conjugate base, conjugate acid, pKb, pKa, and how they all relate. It's easy to mix up the meanings of these definitions.

Here are the two other pieces of information you need to know to correctly answer this question:

Strong acids have a weak conjugate base.
Strong bases have a weak conjugate acid.

So if the problem says you have a strong conjugate base, then the molecule must be a weak acid. To illustrate this, think of ammonium, NH4+. Ammonium is a weak acid, but the conjugate base of ammonium is ammonia, NH3, which is a reasonably good base.

Learn more about conjugate base here : brainly.com/question/22514615

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When balancing a chemical equation, the number of h atoms in 2 ch4 is eight?

Viktor [21]

Yes... that is correct.

CH4 is methane so the coefficent in front of it would double the number of atoms of each element

6 0

3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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3 years ago

0.00099989<br> 3 sig figs, scientific notation

mixas84 [53]

1.00*10^3
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zalisa [80]

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