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anastassius [24]
3 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]3 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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Consider the reaction: P4 + 6Cl2 = 4PCl3.
likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

                                P4      +      6Cl2      =      4PCl3

                          4(31) ---------- 12(35.5)

                         20     ----------    x

                    x = 20(12x35.5) / 4(31)

                   x = 8520 / 124

                   x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

                            P4      +      6Cl2      =      4PCl3

                       124g             426 g               4(31 + 3(35.5)) = 550g

                        15g               22g

I will use P4 to find the limiting reactant

                 

                     x = (15 x 426) / 124 = 51.5   The limiting reactant is Chlorine

                                                                  because we need 51.5 g and we only have 22g

Excess reactant

                 x = (22 x 124) / 426 = 6.4 g of P4

           Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

                              426 g of Cl2 ----------------  550 g of PCl3

                                 22g of Cl2 ------------- -     x

            x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

   % yield = (16.25  - 28.4) / 28.4 x 100

  % yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

                                   124g of P4  -------------      426 g  6Cl2

                                     28g           ---------------     x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3  - 96.2 = 10.1 g of Cl2 in excess

                   

                 

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Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

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