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babunello [35]
3 years ago
15

How do carbon 12 and carbon 13 differ?

Physics
1 answer:
astraxan [27]3 years ago
4 0

Answer:

A) & B)

Explanation:

First, the numbers 12 and 13 represent the <em>atomic mass number</em> of the atoms.

  • So, A) is true: thus Carbon12 and Carbon 13 have different mass numbers.

The <em>mass number</em> is equal to the total number of protons and neutrons. Consider that any element has the same number of protons, regardless of the number of neutrons. The number of protons in Carbon is 6.

The amount of neutrons can be calculated by: <em>mass number </em>minus <em>number of protons.</em>

For Carbon 12: 12-6=6

For Carbon 13: 13-6=7

  • B) is true, we just proved they have different amounts of neutrons.

In order for the charge of the atom to be neutral, the amount of electrons must be equal to the number of protons (as they have opposite charges). And we now know that the number of protons in Carbon12 and Carbon13 are always the same

  • C) is false, the number of electrons is the same in both atoms

The atomic charges in both are neutral, due to the fact that they have the same amount of protons and electrons in both cases. Is only the neutrons (thus the mass numbers)that change

  • D) is false, they have equal atomic charges
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Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

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3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

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zysi [14]

Answer:

106.24 kJ.

Explanation:

Given that,

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