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Gre4nikov [31]
3 years ago
11

When Dr. Montero was observing the endoplasmic reticulum of a cell using an electron microscope, she noticed that it was covered

with ribosomes. What structure was she observing?
A
smooth endoplasmic reticulum

B
rough endoplasmic reticulum

C
convoluted endoplasmic reticulum

D
porous endoplasmic reticulum
Physics
1 answer:
Vesnalui [34]3 years ago
7 0
The answer is Convoluted endoplasmic reticulum
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Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain
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Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

The initial mass of the sample = 1 microgram

After 5 half-lives, the mass should reduce to 1/32 of the original.

So the concentration left = 1/32 of 1 microgram = 1/32 microgram

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3 years ago
Which of the following represents an element?
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What is the primary force that causes the seafloor to spread and continents to drift?
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5 0
4 years ago
A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one ele
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Answer:

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Liquid water at 25°C and 1 bar fills a rigid vessel. If heat is added to the water until its temperature reaches 50°C, what pres
Tcecarenko [31]

Answer:

The developed pressure is 205.75 bar.

Explanation:

Given that,

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Value of \beta = 36.2\times10^{-5}\ K^{-1}

Value of k=4.42\times10^{-5}\ bar^{-1}

Specific volume of liquid = 1.0030 cm³/g

We need to calculate the final pressure

Using formula for constant volume change

\beta(T_{2}-T)_{1})-k(P_{2}-P_{1})=0

Put the value into the formula

36.2\times10^{-5}(323-298)-4.42\times10^{-5}(P_{2}-1)=0

P_{2}=\dfrac{36.2\times10^{-5}(323-298)}{4.42\times10^{-5}}+1

P_{2}=205.75\ bar

Hence, The developed pressure is 205.75 bar.

5 0
3 years ago
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