Answer:
Explanation:
The first step is the <u>calculation of the moles</u> of 
and 
, so:
Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:
Now we can <u>convert the grams</u> of O to moles, so:
The next step is to divide all the mol values by the <u>smallest one</u>:
Therefore the formula is
The gas will obey Boyles Law:
P1 V1 = P2 V2 where P1 and V1 are the original pressure and volume and P2 and V2 are the new values.
If V2 = 2V1 (given) then:
P1 V1 = P2 *2 V1
P2 = P1 V1 / 2V1
P2 = P1 / 2
In other words the pressure is halved. (answer).
It indicates the number of moles of reactants and products
Explanation:
The coefficients in front of the reactants and products in a chemical reaction represents the number of moles of reactants and products.
Every reaction is made up of equal number of moles of reactants and products. Thus, chemical equations are written in such a way to obey the law of conservation of matter.
The numbers used are usually whole numbers and the are very important in stoichiometry.
learn more:
Number of moles brainly.com/question/1841136
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The required mole ratio of NH₃ to N₂ in the given chemical reaction is 2:4.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the number of entities present on the reaction before and after the reaction.
Given chemical reaction is:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
From the stoichiometry of the reaction it is clear that:
4 moles of NH₃ = produces 2 moles of N₂
Mole ratio NH₃ to N₂ is 2:4.
Hence required mole ratio is 2:4.
To know more about mole ratio, visit the below link:
brainly.com/question/504601
<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
