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anastassius [24]
2 years ago
5

The first eighteen chemical elements on the periodic table are described below in a series of statements. The first 1S letters o

f the alphabet (A-R) have been arbitrarily assigned to replace the standard chemical symbols for these elements.
From the information given below, determine the proper location of the "mystery" elements.

1) The following elements belong together in groups: BF, EQ, PR, DC, J0, NK, G& HIM
2) Element D has four valence electrons.
3) Element E has the greatest electronegativity in its period.
4) Element B has fairly low first and second ionization energies and a lower boiling point than the other member of its group.
5) Concerning chemical reactivity in one period, the following are in increasing order: MFL
6) Element A really doesn't have a family to call its own and is a diatomic gas at room
7) Element I will not combine with P or Q
8) Element G is larger than M
9) The compound that forms between R and K ions has the formula R2K3.
10) Element R is more metallic than its family partner
11) At room temperature, Element O is a diatomic gas, while J is a nonmetallic solid.
12) Element Qis heavier than M, but slightly larger than MM
13) When Element C loses its valence electrons, it has the same electron structure as neutral H
14) Element I has the highest nuclear charge of the elements listed.
15) N's first ionization energy is greater than K's
Chemistry
1 answer:
Aleonysh [2.5K]2 years ago
3 0

Answer: The questions looks unclear

Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.

1. D belongs to group 4

E. Belongs to group 7

B belongs to group 1

A belongs to group 8. A noble gas.

R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8

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Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
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Your answer is right.

Important elements to consider:

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