An amoeba has 1.00×1016 protons and a net charge of 0.300 pC.

 How many fewer electrons are there than protons?

 If you paired them up, what fraction of the protons would have no electrons?

We need to solve this problem in two parts: 

1. **Determine the number of fewer electrons than protons**:  
The amoeba has \(1.00 \times 10^{16}\) protons, and the net charge of the amoeba is 0.300 pC (picocoulombs).  
We know that the charge of a single proton is \(1.60 \times 10^{-19}\) C.  
The number of excess protons can be calculated by dividing the total charge by the charge of a single proton.

2. **Determine the fraction of protons that have no paired electrons**:  
This fraction will be equal to the ratio of excess protons (those not paired with electrons) to the total number of protons.

### Step 1: Number of Excess Protons

The net charge of the amoeba is 0.300 pC. Converting this to coulombs:
\[
0.300 \text{ pC} = 0.300 \times 10^{-12} \text{ C}
\]
To find the number of excess protons:
\[
\text{Number of excess protons} = \frac{\text{Net Charge}}{\text{Charge of a single proton}} = \frac{0.300 \times 10^{-12} \text{ C}}{1.60 \times 10^{-19} \text{ C/proton}}
\]

Let's calculate that first.The number of excess protons is 1,875,000. This means there are 1,875,000 fewer electrons than protons.

### Step 2: Fraction of Protons Without Paired Electrons

Now, we calculate the fraction of protons that do not have paired electrons. This fraction is the ratio of the number of excess protons to the total number of protons:

\[
\text{Fraction} = \frac{\text{Number of Excess Protons}}{\text{Total Number of Protons}} = \frac{1,875,000}{1.00 \times 10^{16}}
\]

Let's calculate that.The fraction of protons that do not have paired electrons is \(1.875 \times 10^{-10}\). 

This means that an extremely small fraction of the protons are without paired electrons, indicating that almost all protons have electrons paired with them.

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Would you like more details or have any questions about the calculations?

Here are 5 related questions for further exploration:

1. What is the total number of electrons in the amoeba?
2. How would the fraction of unpaired protons change if the net charge increased?
3. What would happen to the net charge if the amoeba had more electrons than protons?
4. How does the charge of an electron compare to that of a proton?
5. How would you calculate the number of neutrons if the mass of the amoeba were given?

**Tip:** In most physics problems involving charges, it’s important to understand the fundamental charge of particles, as this helps in converting between charge and the number of particles.

Learn how to calculate the number of excess protons compared to electrons in an amoeba with a net charge of 0.300 pC. This problem explores electrostatics, charges, and the relationship between protons and electrons. Suitable for high school students.

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