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weeeeeb [17]
3 years ago
5

Iron and chlorine form an ionic compound whose formula is FeCl3. The name of this compound is __________. Group of answer choice

s iron chlorine iron (III) chloride moniron trichloride iron (III) trichloride ferric trichloride
Chemistry
1 answer:
il63 [147K]3 years ago
8 0

Answer:

Iron and chlorine form an ionic compound whose formula is FeCl3. The name of this compound is iron (III) chloride or iron trichloride.

Explanation:

Iron (III) chloride or Iron trichloride is also known as ferric chloride, it is a common compound of iron in the +3 oxidation state. The anhydrous compound is a crystalline solid with a melting point of 307.6 °C.

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What is the molarity of a 5.00x10^2 mL solution containing 21.1 g of potassium bromide (KBr)? The molar mass of KBr is 119.0 g/m
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Volume of solution in liters:

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number of moles:

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21.1 / 119.0 => 0.1773 moles

Molarity = number of moles / volume 

M = 0.1773 / 0.5

M = 0.355 mol/L

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What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
  • M acid= 0.129 M
  • V base= ?
  • M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>

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3 years ago
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