CHEMISTRY HIGH SCHOOL

how much oxygen would be produced by heating 20 g of mercury (II) oxide if 18.6 g of mercury is produced?

Answers

Answer 1
Answer:

According to the stoichiometry of the reaction of mercury(II) oxide 1.477 g of  oxygen would be produced by heating 20 g of mercury (II) oxide if 18.6 g of mercury is produced.

What is stoichiometry?

Stoichiometry is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample.

433.18 g mercury (II) oxide gives 32 g oxygen thus, 20 g  of mercury (II) oxide will produce 20×32/433.18= 1.477 g of oxygen.

Thus, 1.477 g of  oxygen would be produced by heating 20 g of mercury (II) oxide if 18.6 g of mercury is produced.

Learn more about stoichiometry,here:

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Answer 2
Answer: Hg2O -> O2 + Hg
2Hg2O -> O2 + 2Hg
20g-> ?g + 18.6g
                     (1 molHg2O )       (1 mol O2)           (32g O2)
 20g 2Hg2O x -------------------   x -------------------  x  -------------- = 0.77g O2
                    (834.36Hg2O)    (1 mol Hg2O)       (1 mol O2)

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Which of the following is a secondary pollutant?A) sulfur dioxide
B) radon
C) carbon monoxide
D) sulfuric acid

Answers

Final answer:

A secondary pollutant is one that is not directly emitted but rather forms when primary pollutants react in the atmosphere. Among the options provided, sulfuric acid is a secondary pollutant. Correct option is D.

Explanation:

In the context of environmental science, a secondary pollutant is a type of pollutant that is not directly emitted. Instead, secondary pollutants are formed when primary pollutants undergo chemical reactions in the atmosphere. Among the options provided, option D) sulfuric acid is a secondary pollutant. The others - sulfur dioxide, radon and carbon monoxide - are primary pollutants, as they're emitted in this form from a source.

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The normal pH of surface sea water is _____. 5.0 6.0 7.0 9.0

Answers

Answer: Option (d) is the correct answer.

Explanation:

It is known that sea water contains a large amount of common salt, that is, sodium chloride (NaCl). This salt is basic in nature.

Since, it is also known that species which have pH equals to 7 are basic in nature. Species which have pH less than 7 are acidic in nature and species which have pH greater than 7 are basic in nature.

Hence, due to the presence of common salt in sea water its surface is basic in nature.

Therefore, we can conclude that the normal pH of surface sea water is 9.0.
The PH level of the surface of the water is 9.0 
Hope this helps 

The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What is the mass, in grams, of H2 that must have reacted, to the correct number of significant figures?

Answers

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

 

We are given the amount of ammonia formed from the reaction. This is where we start our calculations.

 

0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2

Answer:

Answer is 0.102.    

Explanation:

The rate law for a hypothetical reaction is rate = k [A]2. When the concentration is 0.10 moles/liter, the rate is 2.7 × 10-5 M*s-1. What is the value of k?

Answers

Answer:The value of rate constant is 2.7* 10^(-3) Ls^(-1) mol^(-1).

Explanation:

Concentration of [A]=0.10 mol/L

The rate of the reaction = 2.7* 10^(-5) M/s

Rate constant = k

The given rate law:

R=k[A]^2

2.7* 10^(-5) M/s=k* (0.10 mol/L)^2

k=(2.7* 10^(-5) mol/L s)/(0.10 mol/L* 0.10 mol/L)=2.7* 10^(-3) Ls^(-1) mol^(-1)

The value of rate constant is 2.7* 10^(-3) Ls^(-1) mol^(-1).

Given:

rate = k [A]2

concentration is 0.10 moles/liter

rate is 2.7 × 10-5 M*s-1

Required:

Value of k

Solution:

rate = k [A]2

2.7 × 10-5 M*s-1 = k (0.10 moles/liter)^2

k = 2.7 x 10^-3 liter per mole per second

Which of the following examples depict(s) a physical property?

Answers

Answer:

\boxed {\tt D. \ All \ of \ the \ above }

Explanation:

A physical property is a property that can be observed without changing the composition.

Some examples of physical properties are:

  • Color
  • Density
  • Length
  • Mass
  • Smell
  • Temperature
  • Boiling Point
  • Hardness

Now, let's examine our answer choices.

Choice A: Sugar is white in color

  • The property is color, which is a physical property.

Choice B: Iron is more dense than aluminum

  • The property is density, which is a physical property.

Choice C: Diamond is harder than graphite.

  • The property is hardness, which is a physical property.

Since all three choices depict physical properties, the best choice is D. All of the above.

Answer:

D

Explanation:

All explain physical no chemical examples were given

An ion is an atom ____.

Answers

It is an atom CHARGED. 
I hope I helped! =D
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