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sveticcg [70]
3 years ago
9

Name an element in the fourth period of the periodic table with a total of 4 4p electrons

Chemistry
2 answers:
Makovka662 [10]3 years ago
5 0
For you to find the correct element, begin by looking only at the fourth row of elements on the periodic table, as indicated by the 4 in front of the 4p. This number tells you the level, or shell, in which these electrons will reside.

The “p” block is the column containing six individual columns of elements on the right most side of the table. In this configuration, elements can have from one “p” electron, up to six.

The question is asking for the element containing 4 “p” electrons, so on the fourth row of the “p” block, count out each element until you reach four “4p” electrons.

4p1-Gallium
4p2-Germanium
4p3-Arsenic
4p4-Selenium

Selenium, atomic number 34, is the element in the fourth row containing four “p” electrons.
kifflom [539]3 years ago
3 0
Beryllium has 4 electrons
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Vladimir79 [104]
What are the statements?.
8 0
3 years ago
Read 2 more answers
What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

5 0
3 years ago
Vitamin C, also known as ascorbic acid, is composed of 40.9 percent carbon, 4.6 percent hydrogen, and 54.5 percent oxygen. The m
Vladimir79 [104]

Answer:

Molecular formula is : C₆H₈O₆  

Explanation:

40.9 % of C, 4.6 % of H, 54.5 % of O means that in 100 g of ascorbic acid, we have 40.9 g of C, 4.6 g of H and 54.5 g of O.

If 1 mol of ascorbic acid weighs 176 g, let's determine the rules of three to know the grams of each element:

100 g of compound has ___ 40.9 g C __ 4.6 g H ___ 54.5 g O

176 g compound mus have:

(176 . 40.9) / 100 = 72 g of C

(176 . 4.6) / 100 = 8 g of H

(176 . 54.5) / 100 = 96 g of O

If we convert the mass to moles:

72 g . 1 mol / 12 g = 6 C

8 g . 1 mol/ 1g = 8 H

96 g . 1mol / 16 g = 6 O

Molecular formula is : C₆H₈O₆  

In conclussion, 1 mol of ascorbic acid has 6 moles of C, 8 moles of H and 6 moles of O

6 0
4 years ago
Help
Burka [1]
The Moon<span> moves </span>around<span> the </span>Earth in<span> an </span>approximately<span> circular </span>orbit<span>, going once </span>around<span>  </span>in approximately<span> 27.3 </span><span>days.  </span>The moon orbits quite fast: it moves about 0.5 degrees<span> per hour in the sky. So, its average movement in a day is around 13 degrees.   5 days x 13 degrees is approximately 65 degrees. </span>
4 0
3 years ago
A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t
lidiya [134]

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

We are given:

K_{sp} of CuCl = 1.0\times 10^{-6}

K_{sp} of CuI = 5.1\times 10^{-12}

Concentration of Cl^-\text{ ion}=0.021M

Concentration of I^-\text{ ion}=0.017M

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

  • <u>For CuCl:</u>

K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M

Concentration of copper (I) ion = 4.76\times 10^{-5}M

  • <u>For CuI:</u>

K_{sp}=[Cu^+][I^-]

Putting values in above equation, we get:

5.1\times 10^{-12}=[Cu^+]\times 0.017

[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M

Concentration of copper (I) ion = 3.00\times 10^{-10}M

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0
4 years ago
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