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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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There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore
devlian [24]

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore,  which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\

Where the short tons are cancelled out as well as the grams, in order to obtain:

0.103 otz

Learn more about dimensional analysis: brainly.com/question/10874167

4 0
2 years ago
Which of the following correctly represents the transmutation in which a curium-242 nucleus is bombarded with an alpha particle
Alekssandra [29.7K]

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.

The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:

_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}

The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.

6 0
3 years ago
What is 0.00550 g converted to mg
Mila [183]

Answer: it is 5.5 mg

Explanation:

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? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2
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Data: 
M_{concentrated} = 2.5\:mol
V_{concentrated} = ?
M_{dilute} = 0.50\:mol
V_{dilute} = 100\:mL\to0.100\:L
<span>
Formula: Dilution Calculations

</span>M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
<span>
Solving:

</span>
M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
2.5 * V_{concentrated} = 0.50 * 0.100
2.5V_{concentrated} = 0.05
V_{concentrated} =  \frac{0.05}{2.5}
\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark<span>







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Explanation:

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