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GarryVolchara [31]
1 year ago
5

A sample of a certain lead compound contains 12.92 g of lead for 2 g of oxygen. A second sample has mass of 34.27 g and contains

14.39 g of oxygen. Are the two compound the same
Chemistry
1 answer:
Elis [28]1 year ago
5 0

The two chemical compounds are not the same, because their ratio is not equal. In both samples the composition of lead and oxygen is different.

<h3>What is a chemical compound?</h3>

A chemical compound is a substance made of numerous similar molecules (or molecular entities) joined by chemical bonds and comprising atoms from various chemical elements. Therefore, a molecule made up of only one type of atom is not a compound. Chemical reactions, which may entail interactions with other molecules, can change a compound into a distinct substance. Atomic bonds may be broken or new ones created during this process.

<h3>What are the calculations?</h3>

sample 1 = mass of lead / mass of oxygen = 12.92g/2g = 6.46 .

sample 2 = mass of lead/ mass of oxygen = 34.27 - 14.39/14.39 = 1.38 .

so, the ratios are not the same.

Hence, the two chemical compounds are not the same, because their ratio is not equal. In both samples the composition of lead and oxygen is different.

To know more about Chemical compounds, check out:

brainly.com/question/26487468

#SPJ1

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The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e
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Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           N_{2}O_{4} \rightarrow 2NO_{2}

Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

Initial :               x                   0

Change :       - 0.1        2 \times 0.1

Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

     K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

                 2 = \frac{(0.2)^{2}}{(x - 0.1)}

                2 \times (x - 0.1) = (0.2)^{2}

                            x = 0.12

This means that P_{N_{2}O_{4}} = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

6 0
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The process in which an organic acid and an alcohol react to form an ester and water is known as esterification. Ethyl butanoate
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Answer:

697 g

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C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O

The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:

6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅

The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:

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7 0
3 years ago
Does a metal conduct electricity in the solid state?
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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
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Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

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3 years ago
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