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sladkih [1.3K]
3 years ago
5

there about 5000 elements, which combine to form the vast number of diffrent substances in the world around us

Chemistry
1 answer:
marissa [1.9K]3 years ago
5 0

Answer: false 100

Explanation:

if this is the true or false test here

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As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit
jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced  0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of  NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of  NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl  = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of  K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of  K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles =  0.0003663 moles

Mass of  K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

7 0
3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

#SPJ4

 

5 0
1 year ago
I need help please my teaching is going to go crazy if I didn’t do it
victus00 [196]

Answer:

No One is going to answer all this because it hard to type it all in her. its easy to search it up

Explanation:

4 0
3 years ago
The formula for lithium oxide is A) Li2O B) LiO2 C) LiO D) Li2O3
Greeley [361]
Li2O is the formula for <span> lithium oxide</span>
4 0
3 years ago
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The reactants of two chemical equations are listed.
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The correct answer is b
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