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Morgarella [4.7K]
1 year ago
13

An unknown metal cylinder has a mass of 34.3 g and a volume of 2.1 mL . What is the density of the metal in g / m * L ?​

Chemistry
1 answer:
saul85 [17]1 year ago
6 0

Taking into account the definition of density, the density of the metal is 16.33 \frac{g}{mL}.

<h3>Definition of density</h3>

Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Density of the metal in this case</h3>

In this case, you know that:

  • Mass= 34.3 g
  • Volume= 2.1 mL

Replacing in the definition of density:

density=\frac{34.3 g}{2.1 mL}

Solving:

<u><em>density= 16.33 </em></u>\frac{g}{mL}

In summary, the density of the metal is 16.33 \frac{g}{mL}.

Learn more about density:

<u>brainly.com/question/952755?referrer=searchResults</u>

<u>brainly.com/question/1462554?referrer=searchResults</u>

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Answer:

  • <u><em>a. C₂H₄</em></u>

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

<u>a. C₂H₄:</u>

  • C₂H₄ (g) + 3O₂ (g) → 2CO₂(g)  + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

  • 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g)  + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

  • C₃H₈ (g) + 5O₂ (g) → 3CO₂(g)  + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

  • 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g)  + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

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For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has
exis [7]

Answer:

70 mL of 5% HCl and 30 mL of 15% HCl

Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

y = 1 - x

This expression is substituted into the first equation and we solve for x.

5x + 15(1 - x) = 8

5x+ 15 - 15x = 8

-10x = -7

x = 7/10 = 0.7

We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

Similarly, the volume of 15% HCl we need is:

(100 mL)(0.3) = 30 mL

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Answer:

c. H2(g) + CuO(s) → Cu(s) + H2O(l)

e. H2(g) + Cl2(g) → 2HCl(g)

Explanation:

A redox reaction is a reaction that involves a changes in oxidation number of the species involved in the reaction.

The oxidizing agent experiences a decrease in oxidation number while the reducing agent experiences an increase in oxidation number.

For H2(g) + CuO(s) → Cu(s) + H2O(l)

Copper is reduced from +2 to 0 while hydrogen is oxidized from 0 to +2 Hence hydrogen is the reducing agent while copper is the oxidizing agent.

For H2(g) + Cl2(g) → 2HCl(g)

Chlorine is reduced from 0 to -1 while hydrogen is oxidized from 0 to +1. Hence chlorine is the oxidizing agent while hydrogen is the reducing agent.

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