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Serga [27]
3 years ago
6

If atoms of a halogen nonmetal (Group 17) gains one electron, the atoms the have __.

Chemistry
2 answers:
RUDIKE [14]3 years ago
8 0

Hm, this could be more than one option, but gaining electrons makes a negative charge, so

If atoms of a halogen nonmetal (Group 17) gains one electron, the atoms the have "a negative one charge".

Lady bird [3.3K]3 years ago
4 0

Answer : If atoms of a halogen nonmetal (Group 17) gains one electron, the atoms the have (-1) (or, negative) change.

Explanation :

Periods : A row in the periodic table is known as periods.

The common features about the elements of a period are :

Each periods has elements with the same number of electron shells or energy levels or shell.

They transition from metal to noble gas that means they move from metal to non-metal and then non-metal to noble gas.

Groups : A column in the periodic table is known as groups.

The common features about the elements of a same group are :

Each groups has elements with the same number of valance electrons.

The elements in the same group have similar chemical properties.

The elements in the same group have similar physical properties.

The group 17 is a non-metal and known as halogen. The number of valance electrons present in group 17 are 7 valence electrons.

The general electronic configuration of group 17 is:

ns^2np^5

As per question, when atoms of halogen gains one electron then the  atoms have (-1) (or, negative) charge and attain stable electronic configuration as noble gas.

For example : The element chlorine belongs to group 17 and the electronic configuration is:

1s^22s^22p^63s^23p^5

When chlorine atom gain one electron then the chlorine atom have (-1) charge Cl^-.

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How many grams of acetylene are produced by adding water to 5.00 g CaC2?
Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

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H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

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5 0
3 years ago
How many moles of ethanol are produced starting with 500.g glucose?
Monica [59]
<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

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

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

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

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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