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2 years ago

A single cell undergoes mitosis every five minutes. How many cells will result from this cell in 15 minutes? In 30 minutes?

Chemistry

2 answers:

Ivanshal [37]2 years ago

7 0

Answer: 15 minutes 3 cells will be reproduced and in 30 minutes 6 cells will be reproduced

Explanation:

Finger [1]2 years ago

6 0

In 15 minutes 3 cells will be reproduced and in 30 minutes 6 cells will be reproduced

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Which of the following phrases describes valence electrons?

kykrilka [37]

Answer:

Explanation:

because valence electrons are located at the last energy level

7 0

2 years ago

Air is transferred from a 75 L tank where the pressure is 125 psi and the temperture is 288 k to a tire with a volume of 6.1 L a

aniked [119]

<h3><u>Answer</u>;</h3>

= 4.68 K

<h3><u>Explanation</u>;</h3>

According to the combined gas law;

P1V1/T1 = P2V2/T2

Given; P1 = 125 Psi

V1 = 75 L

T1 = 288 K

P2 = 25 PSI

V2 =6.1 L

Therefore;

T2 = P2V2T1/P1V1

= (25×6.1 ×288)/(125×75)

= 4.6848

= 4.68 K

4 0

2 years ago

The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what wi

Kisachek [45]

Answer:

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^(-0.0546*4)

A = 845 e^(-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours

7 0

2 years ago

A tank containing both hf and hbr gases developed a leak. The ratio of the rate of effusion of hf to the rate of effusion of hbr

Pavlova-9 [17]

Answer:

2.01

Explanation:

The effusion is the passage of the molecules by a small hole by a difference of pressure. By Graham's Law, the rate of the effusion is inversely proportional to the square of the molar mass of the compound. Thus,

rateHF/rateHBr = √MHBr /√MHF

MHBr = 81 g/mol

MHF = 20 g/mol

rateHF/rateHBr = √81/√20

rateHF/rateHBr = 2.01

4 0

2 years ago

Automobiles are often implicated as contributors to global warming because they are a source of the greenhouse gas CO2. How many

kondor19780726 [428]

Answer:

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

Explanation:

Mileage of the car = 27.5 miles/gal

Distance driven by car in week = 170 miles

Distance driven by car in a day= $\frac{170 miles}{7}=24.286 mile$

Volume of gasoline used in a day : V

$\frac{24.286 mile}{27.5 miles/gal}=0.8831 gal=0.8831\times 3785.41 cm^3=3,342.96 cm^3$

$1 gal = 3785.41 mL = 3785.41 cm^3$

Density of the gasoline = d= $0.692 g/cm^3$

Mass of the gasoline used in a day = m

$m=d\times V=0.692 g/cm^3\times 3,342.96 cm^3=2,313.33 g$

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

Moles of gasoline ( isooctane):

= $\frac{2,313.33 g}{114 g/mol}=20.292 mol$

According to reaction , 2 moles of isooctane gives 16 moles of carbon dioxide gas.Then 20.292 mole isooctane will give :

$\frac{16}{2}\times 20.292 mol=162.340 mol$ of carbon dioxide gas

Mass of 162.340 moles of carbon dioxide gas :

162.340 mol × 44 g/mol = 7,142.91 g

Mass of carbon dioxide gas produced in a day = 7,142.91 g

1 year = 365 days

Mass of carbon dioxide gas produced in a 365 days:

= 365 × 7,142.91 g = 2,607,163.494 g

1 pound = 453.592 g

$2,607,163.494 g=\frac{ 2,607,163.494}{453.592} pounds=5,747.82 pounds$

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

7 0

2 years ago