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Irina18 [472]
3 years ago
13

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has i

ncreased to 570. (a) Find an expression for the number of bacteria after t hours. P(t)
Biology
1 answer:
Naddika [18.5K]3 years ago
5 0

Answer:

p(t)=100e^ln5.7t

Explanation:

The growth rate of bacteria is directly proportional to its size and shows the exponential growth. The initial bacteria amount is 100 and later increased to 570.

The number of bacteria after t hours is

p(t)=100e^kt where k is the relative growth rate. and t is the time.

Here p (t) = 570 and t = 1 hour, then K =

 420=100e^k(1)  

K = ln(5.7)

Put the value of K in original equation then,

p(t)=100e^ln(5.7)t

Thus, the expression for the number of bacteria after t hours is p(t)=100e^ln(5.7)t .

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I couldn’t fit the whole question in so the last two options are
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A certain buffer is made by mixing a weak acid HA and its conjugate base A–. When HA is present at a concentration of 0.5 mM and
MAVERICK [17]

Answer:The ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

Explanation:

The dissociation constant for formic acid = K_a=1.8\times 10^{-4}

Concentration of HA= 0.5 mM

Concentration of A^-= 0.1 mM

pH = 6.16

First we have to calculate the value of K_a.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[A^-}{[HA]}

Now put all the given values in this expression, we get:

6.16=pK_a+\log (\frac{[0.1]}{0.5})

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To calculate the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02.

Using Henderson Hesselbach equation :

7.02=6.86+\log \frac{[A^-]}{[HA]}

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\frac{[A^-]}{[HA]}=1.44

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Thus the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

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