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Katena32 [7]
3 years ago
7

Which star is hotter,but less luminous, than polaris

Chemistry
1 answer:
steposvetlana [31]3 years ago
6 0
It would be any white dwarf star
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You are playing a game “will it float?” In this game, you are given a large, square can of tuna. If you know the density of wate
saul85 [17]

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I think it's just density of the tuna

5 0
3 years ago
Is trimming a bush because it grows too tall a chemical or physical change
professor190 [17]

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Physical Change

Explanation:

The bush is changing shapes, not changing what it is.

8 0
4 years ago
you are running a transformation reaction, and in your microfuge tube currently is your plasmid and your insert fragments (both
ElenaW [278]

A foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction.

<h3>How to explain the reaction?</h3>

With the aid of two enzymes, ligase and restriction enzymes, a foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction. Each enzyme detects a target DNA sequence and cuts it nearby, while ligase aids in connecting the DNA. When two bits of DNA have complimentary bases, it facilitates their joining.

Plasmid and the insert fragment are both present in the microfuge tube, and they both have compatible sticky ends. However, the ligase has been denatured and is no longer active because the prior student left it outside rather than freezing it; despite this, we had already put the ligase into the tube. Ligase aids in binding the plasmid and insert fragments together, but because it is denatured in this instance, it will no longer be able to do so. As a result, no transformation process will take place. And since ligase links DNA fragments together by catalyzing the development of connections between the nearby nucleotides, the two fragments will not be able to unite.

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4 0
2 years ago
[HCN]=0.09974 M<br> Kp=7.52<br> Calculate the partial pressure of HCN ?
kvv77 [185]

Answer:

i don't understand

Explanation:

6 0
3 years ago
How much of a sample remains after five half-lives have occurred?
timama [110]

Answer:

1/32 of the original sample

Explanation:

We have to use the formula

N/No = (1/2)^t/t1/2

N= amount of radioactive sample left after n number of half lives

No= original amount of radioactive sample present

t= time taken for the amount of radioactive samples to reduce to N

t1/2= half-life of the radioactive sample

We have been told that t= five half lives. This implies that t= 5(t1/2)

N/No = (1/2)^5(t1/2)/t1/2

Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;

N/No= (1/2)^5

N/No = 1/32

Hence the fraction left is 1/32 of the original sample.

3 0
3 years ago
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