MHB Webpage title: Calculating Coulombs in a Nickel Coin

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A nickel coin weighing 5 grams contains approximately 2.3 x 10^5 coulombs of charge. The calculation assumes the coin is made of nickel, which has an atomic number of 28 and an atomic mass of 58.7 u. By determining the number of moles in the coin and multiplying by Avogadro's number, the total number of electrons is calculated. Each nickel atom contributes 28 electrons, leading to the final charge value. The discussion confirms the assumption that the coin is indeed made of nickel.
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How many Coulombs of charge positive and negative are there is a coin of 5 g of mass?
Answe: 2,3.105 C

I don't know but I have no idea I think there is no enough data
 
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Do you know of what metal the coin is made?
 
leprofece said:
How many Coulombs of charge positive and negative are there is a coin of 5 g of mass?
Answe: 2,3.105 C

I don't know but I have no idea I think there is no enough data

If we suppose that the coin is made of nickel, the atomic number is $28$ and the atomic mass is $58.7 \text{ u}$.

Therefore $5 \text{ grams}$ contains $\frac {5}{58.7} \text{ mol}$ in metal atoms.
Multiply by $N_A$ and we find $\frac {5}{58.7} \cdot 6.022\cdot 10^{23}\text{ atoms}$.
With 28 electrons that corresponds to:
$$\frac {5}{58.7} \cdot 6.022\cdot 10^{23} \cdot 28 \text{ electrons}$$
Convert to coulombs and we get:
$$\frac {5}{58.7} \cdot \frac{6.022\cdot 10^{23}}{6.241\cdot 10^{18}}\cdot 28 = 2.3\cdot 10^5 \text{ coulombs}$$

I think the coin is made of nickel! ;)
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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