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Oxana [17]
3 years ago
12

Sodium chloride comprises 97%-99% of table salt. Given their location on the periodic table, identify the ionic charge for each

element and predict the chemical formula of the compound formed. A) Sodium, Na, is an alkali metal and is in Group IA with an ionic charge of +1. While chlorine, Cl, is a halogen and is in Group VIIA with a charge of -1. These will form NaCl(s). B) Sodium, Na, is an alkaline earth metal and is in Group IA with an ionic charge of +2. While chlorine, Cl, is a halogen and is in Group VIIA with a charge of -1. These will form NaCl2(s). C) Sodium, Na, is an alkali earth metal and is in Group IA with an ionic charge of +2. While chlorine, Cl, is a noble gas and is in Group VIA with a charge of -2. These will form Na2Cl2(s). D) Sodium, Na, is an alkali earth metal and is in Group IA with an ionic charge of +1. While chlorine, Cl, is a diatomic gas and is in Group VIA with a charge of -2. These will form Na2Cl(s).
Chemistry
2 answers:
ruslelena [56]3 years ago
7 0
The answer is A) Sodium is an Alkali metal and is in group IA with an ionic charge of +1. While chlorine, Cl, is a halogen and is in group VIIA with a charge of -1. These will form NaCl(s).
lubasha [3.4K]3 years ago
7 0

Answer is: A) Sodium, Na, is an alkali metal and is in Group IA with an ionic charge of +1. While chlorine, Cl, is a halogen and is in Group VIIA with a charge of -1. These will form NaCl(s).

Table salt is sodium chloride mixed with small amount of potassium iodide (KI), sodium iodide (NaI) or sodium iodate (NaIO₃).

Atomic level - sodium chloride (NaCl) has crystal cubic structure (lattice-type arrangement) with ionic bonds. Sodium is cation with charge 1+ and chlorine is an anion with charge 1-.

Macroscopic level - table salt is colorless crystal, soluble in water with high melting and boiling temperature.

Ionic compounds are good good electricity and heat conductors, because ionic compounds have mobile ions (cations and anions) that are able to transfer electrical charge.

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That's kind of a ponderous way to describe it, but your 'X' represents
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3 0
3 years ago
Divide the numbers in around the answer to the correct number of significant figures.
Zina [86]

Answer:

\large \boxed{19.0}

Explanation:

In multiplication and division problems, the answer can have no more significant figures than the number with the fewest significant figures.

A calculator gives the result:

\dfrac{163.8}{8.64} = \mathbf{18.95833333}

163.8    has four significant figures.

   8.64 has three significant figures.

You must round to three significant figures.

That is, you drop all the digits to the right of the 9 — the red line in Fig. 1 below. You are rounding to the nearest tenth.

To round a number to the nearest tenth, you look at the digit in the hundredths place (7). See Fig. 2.

If the digit to be dropped is greater than 5 or is 5 followed by at least one non-zero digit, you add 1 to the number in the tenths place (Fig. 3).

Here's how you do it.

  • The digit in the hundredths place is 5 followed by non-zero digits.
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  • ⁹/₁₀ + ⅒ = ¹⁰/₁₀ = 1 + ⁰/₁₀ = 1.0

The tenths digit becomes 0 and the ones digit increases from 8 to 9.

\text{The quotient of $\dfrac{163.8}{18.64}$ is $\large \boxed{\mathbf{19.0}}$}

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1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →
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The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

  • The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

    If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

    But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.

  • 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

    As 80 g NaOH produces 106 g of Na₂CO₃.

    Therefore 5 g NaoH will produce ;

    106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

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