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dybincka [34]
3 years ago
15

If you know chemistry please be so kind and take some time to help me with these 2 problems!! Marking brainliest!!

Chemistry
1 answer:
Nataly_w [17]3 years ago
6 0

Our balanced equation is Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag.

The coefficients of reactants tell us that 1 mole of Cu reacts per 2 moles of AgNO₃. We want to know the mass of Cu that will react with a mass (specifically, 420 g) of AgNO₃. To start, we must convert the mass of AgNO₃ to moles of AgNO₃:

Moles of AgNO₃ = Mass of AgNO₃ ÷ Molar Mass of AgNO₃

Moles of AgNO₃ = 420 g AgNO₃ ÷ 169.88 g AgNO₃/mol AgNO₃

Moles of AgNO₃ = 2.47 moles of AgNO₃.

Since only half as many moles of Cu is consumed in reacting with the AgNO₃, the number of moles of Cu that will react with 2.47 moles of AgNO₃ would be half of 2.47, or 1.24 moles of Cu.

Finally, we can convert the molar quantity of Cu to mass by multiplying the number of moles of Cu by the molar mass of Cu:

Mass of Cu = Moles of Cu × Molar Mass of Cu

Mass of Cu = 1.24 moles Cu × 63.546 g Cu/moles Cu

Mass of Cu = 78.6 g of Cu

So, 78.6 grams of Cu are needed to react with 420 g of AgNO₃.

---

The way to solve the second question is identical to how we solved the first question. The coefficient of Ag is 2, so the molar ratio between the Ag produced and Cu reacted is 2:1. In other words, half as many moles of Ag produced is equivalent to the number of moles of Cu that was consumed in the reaction.

So, as we did with the first question, we first convert the mass of the Ag produced to moles. The molar mass of Ag is 107.8682 g/mol. Dividing 98.5 g of Ag by the molar mass gives us 0.913 moles of Ag produced. The number of moles of Cu that reacted to produce 0.913 moles of Ag would be 0.913/2 = 0.457 moles of Cu.

Lastly, multiplying 0.457 moles of Cu by the molar mass of Cu, we obtain the mass of Cu (in grams) that reacted: 0.457 × 63.546 = 29.0 g of Cu.

So, 29.0 grams of Cu reacted to produce 98.5 grams of Ag.

Note: I have provided the answers for both questions to three significant figures; while 420 technically has only two significant figures, it would seem consistent to treat it as having three since 98.5 has three digits and three significant figures.

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katrin2010 [14]

Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

<em>Melting point</em> is the temperature at which the substace changes its state from<em> solid to liquid</em>. During that change, <em>the temperature</em> of the substance <em>does not change</em>, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means <em>that portion of the diagram is flat</em>.

The same is valid during<em> boiling</em>: the temperature remains constant while the substance is passing<em> from liquid to gas</em> at the boiling point.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Between the <em>melting</em> (-183°C) and<em> boiling</em> (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a <em>diagonal</em> (slant) <em>line</em> between those points. During this interval the heat is used to <em>increase the temperature</em> and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

<u><em></em></u>

The table shows that the<em> boiling point</em> of gold is 2,856°C.

In a <em>temperature-vs.-time diagram</em> the<em> temperature is represented on the vertical axis (y-value)</em> and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during <em>boiling,</em> the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the <em>top horizontal line in the diagram</em>.

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

<u><em></em></u>

The table shows that the <em>melting point </em>of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

<u>In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the </u><em><u>temperature - vs. - time diagrams </u></em><u>show flat lines (constant y-values) during those intervals of time.</u>

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Answer:

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Explanation:

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Explanation:

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Until 1954, 0 °C on the Celsius scale was defined as the melting point of ice and 100 °C was defined as the boiling point of water under a pressure of one standard atmosphere; this close equivalence is taught in schools today. However, the unit “degree Celsius” and the Celsius scale are currently, by international agreement, defined by two different points: absolute zero, and the triple point of specially prepared water. This definition also precisely relates the Celsius scale to the Kelvin scale, which is the SI base unit of temperature (symbol: K). Absolute zero—the temperature at which nothing could be colder and no heat energy remains in a substance—is defined as being precisely 0 K and −273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C.

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