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Mnenie [13.5K]
3 years ago
5

Which of the following pairings usually forms molecular compounds?

Chemistry
1 answer:
nikdorinn [45]3 years ago
7 0

A nonmetal and a nonmetal will make molecular compounds like H2O and CO2

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A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben
Anit [1.1K]

<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

The chemical equation for the reaction of benzoic acid and methanol is:

\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = \frac{1}{1}\times 0.0295=0.108 moles of methyl benzoate

  • Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

  • To calculate the percentage yield of methyl benzoate, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

6 0
3 years ago
2) Convert 2.65*10^ ^ 25 atoms of Chlorine to moles of CI
Irina18 [472]

Answer:

22 mol

Explanation:

Given data:

Number of atoms of Cl = 2.65×10²⁵ atom

Number of moles of Cl = ?

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.

1 mole = 6.022 × 10²³ atoms

2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms

0.44×10² mol

22 mol

7 0
3 years ago
For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13
igor_vitrenko [27]
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
8 0
3 years ago
If you produced 76.10 grams of potassium chloride in the reaction how many grams of potassium were required?
11111nata11111 [884]

Answer:39.8375

Explanation:

The mole for the equation is 1:1

Then the molar mass of KCl is 74.5g

Molar mass of k is 39

74.5g of KCl gives 39g of k

76.10g of KCl gives xg of k

X= 76.10×39/74.5

X= 2967.9/7

X= 39.8375

6 0
3 years ago
Can sedimentation and decantation be used for all types of mixtures? Explain
Bumek [7]

Explanation:

1. Sedimentation and decantation cannot be used for all types of mixtures.

Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.

In decantation, gravity is used to bring the denser materials to settle at the bottom.

For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.

2. Yes, mass of an object reduces the settling time of such object in a mixture.

The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.

4 0
3 years ago
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