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EleoNora [17]
4 years ago
9

How would a Bohr model of a neon atom differ from the model of nitrogen.

Biology
1 answer:
sasho [114]4 years ago
5 0
"According to the Bohr model an atom has a nucleus with electrons in circular orbit around the nucleus. For a bohr model of Nitrogen there will be 2 orbitals with the first one having 2 electrons and the last orbital having 5. Whereas for the Bohr model of Neon there will be 3 orbitals the first one having 2, the second one having 8 and the last one having 7 electrons."

This answer was by TheMysteriousGamer11, I couldn't have thought of a better way to explain.
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Jayden wants to model the hydrological cycle. He puts some water in a small plastic cup, draws a line on the cup at the water le
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3 years ago
The count in a bacteria culture was 100 after 20 minutes and 2000 after 40 minutes. Assuming the count grows exponentially, What
Afina-wow [57]

Answer:

Explanation:

Let X be the number of bacteria at time t, and Xo be the initial number of bacteria

Bacteria grows exponentially, the exponential growth model is thus:

X = Xo e^{kt}

[<em>e</em> is the exponential sign]

k is the growth constant = growth rate

At t = 20 minutes, X = 100

At t = 40 minutes, X = 2000

Substituting that into the formula

(i) ...    100 = Xo e^{k20}

(ii) ...    2000 = Xoe^{k40}

<em>Divide (ii) by (i)</em>

\frac{2000}{100} = \frac{Xoe^{40k} } {Xoe^{20k}}

20 = e^{40k-20k} = e^{20k}

<em>[Xo cancels Xo. Division of values raised to a power is done by subtracting their powers]</em>

<em>Take the natural log of both sides</em>

㏑20 = 20k

<em>[taking the ㏑ cancels the exponential]</em>

k = \frac{ln20}{20}

<em>We can now substitute k to solve one of the equations</em>

<em>substituting k in (ii)</em>: 2000 = Xo e\frac{ln 20}{20}.^{40}

2000 = Xo e^{0.15*40}

2000 = Xo e^{6}

<em>Making Xo the subject of the formula</em>

Xo = \frac{2000}{e^{6} }

Xo is approximately 5 cells.

THE DOUBLING TIME

The doubling time is the time it takes for the population to double, so 5 cells become 10 cells

Since Xo = 5

Given that X = Xo e^{kt}

When X=10,

10 = 5 e^{0.15t}

<em>solving for t:</em>

e^{0.15t} = \frac{10}{5}

Take ln of both sides

0.15t = ln 2

t =\frac{ln2}{0.15}

t =4.62 minutes

POPULATION AFTER 65 MINUTES

X = 5 e^{0.15*65}

X = 85770

WHEN WILL THE POPULATION REACH 12000

12000 = 5 e^{0.15t}

e^{0.15t} =240

take ln of both sides

0.15t = ln 2400

t = 51.89 minutes (approximately 52 minutes)

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Answer:

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Explanation:

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Answer:

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