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bazaltina [42]
3 years ago
11

The formula is used to calculate the yield of a reaction.

Chemistry
2 answers:
creativ13 [48]3 years ago
6 0

Answer;:

Theoretical yield depends on limiting reagent.

Theoretical yield: no of moles of limiting reagent x molar mass of product

Percentage yield= observed yield x100/ theoretical yield

frosja888 [35]3 years ago
5 0
<span>The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction.</span> <span>The actual yield will never be 100 percent due to limitations.</span>

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How many formula units are in 5.33 moles of cucl2​
Blizzard [7]

Answer:

32.1 × 10²³ formula units of CuCl₂

Explanation:

Given data:

Number of moles of CuCl₂ = 5.33 mol

Number of formula units  of CuCl₂ = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

one mole of  any substance contain 6.022 × 10²³ formula units thus,

5.33 moles of CuCl₂ = 5.33 ×6.022 × 10²³ formula units

32.1 × 10²³ formula units of CuCl₂

3 0
3 years ago
When gas particles collide with a surface
sergij07 [2.7K]
The average kinetic energy of a collection of gas particles depends on the temperature of the gas and nothing else.
3 0
3 years ago
The balanced reaction between aqueous nitric acid and aqueous strontium hydroxide is ________.
Kaylis [27]
<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H                                                  </span>4H
8O                                                  8O
2N                                                  2N
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</span>
3 0
3 years ago
If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
Tresset [83]
<span>atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago</span>
8 0
3 years ago
What is the gram formula of na2co3
makkiz [27]

Answer:

106 gfm

Explanation:

element. number masses

Na. 2 23×2

C. 1. 12

O. 3. 16×3

then add

106 gfm

5 0
3 years ago
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