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viktelen [127]
3 years ago
6

A modern-day nickel is 25% nickel and 75% copper. Every nickel weighs exactly five grams. Using this information, how many nicke

l coins do you need to have before you have a mole of nickel?
Chemistry
1 answer:
Liula [17]3 years ago
7 0

Answer:

  • <u><em>47 coins</em></u>

Explanation:

The atomic mass of <em>nickel</em> is 58.963g/mol. Thus, one mole of nickel weighs "exactly" 58.693g.

How much nickel does every coin contain?

        25\%\text{ of }5g=0.25\times 5g=1.25g\text{ of nickel per coin}

How many times 58.693g contains 1.25g?

         \dfrac{58.693g}{1.25g/coin}=46.95\approx47coins

That means that you need to have 47 coins of nickel to have one mole of nickel.

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Answer:

2.5

Explanation:

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Explain how electron microscopes work and why some scientists might prefer to use electron microscopes instead of light microsco
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Answer:

Electron microscopes differ from light microscopes in that they produce an image of a specimen by using a beam of electrons rather than a beam of light. Electrons have much a shorter wavelength than visible light, and this allows electron microscopes to produce higher-resolution images than standard light microscopes.

Explanation:

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3 years ago
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
A student develops the list shown below that includes laboratory equipment and materials for constructing a voltaic cell.
PSYCHO15rus [73]
Well this answer is not true and is false
3 0
4 years ago
For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

7 0
4 years ago
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