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Elodia [21]
3 years ago
8

Ag, CO, Find out the stock system thank you

Chemistry
1 answer:
Yuri [45]3 years ago
5 0

Answer:

a system in chemical nomenclature and notation of indicating the oxidation state of the significant element in a compound or ion by means of a Roman numeral that is used in parentheses after the name or part of the name designating this element and ending invariably in -ate in the case of an anion and that is placed. Stock nomenclature for inorganic compounds is a widely used system of chemical nomenclature developed by the German chemist Alfred Stock and first published in 1919. In the "Stock system", the oxidation states of some or all of the elements in a compound are indicated in parentheses by Roman numerals.

Explanation:

Brainliest please?

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Carboxylic acid derivatives undergo hydrolysis to make carboxylic acids.
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Answer:

TRUE

Explanation:

All carboxylic acid derivatives have in common the fact that they undergo hydrolysis (a cleav- age reaction with water) to yield carboxylic acids. with hydroxide ion to yield a carboxylate salt and an alcohol. The carboxylic acid itself is formed when a strong acid is subsequently added to the reaction mixture.

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6.0 mol NaOH reacts with
lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • NaOH: 3 moles
  • H₃PO₄: 1 mole
  • H₂O: 3 moles
  • Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

moles of Na_{3}P O_{4} =\frac{6  moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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Calcium carbide reacts with water to produce acetylene gas according to the following equation: CaC2(s) + 2H2O(l)C2H2(g) + Ca(OH
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Answer:

33.7

Explanation:

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