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liq [111]
3 years ago
5

A liquid that evaporates at a slow rate exhibits __________.

Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0
Strong internolecurar forces (A) hope it helps
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Suppose that for the same 10.0 ml sample described in q3, the mass of the crucible with precipitate was 17.550 g, and the mass o
marusya05 [52]
<span>The weight of the precipitate was 17.550g - 17.410g = 0.140g. From a 10ml sample, this means the concentration was 0.140g / 10ml, which is equal to 0.014g/ml.</span>
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In which types of cells must mutations occur in order for the mutation to be passed on to future generations?
Artyom0805 [142]

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The answer is option c which is gametes

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2 years ago
What happens when you mix hydrogine, potassium and sodium
Anika [276]

Answer:

-

Explanation:

As the piece of metal skitters across the surface of the water in a beaker and — particularly in the case of potassium — it appears to catch fire, it is not obvious that the explanation for both phenomena lies in the production of hydrogen gas.

3 0
2 years ago
Which area should have the lowest temperatures?<br><br> A. 1<br><br> B.2 <br><br> C.3 <br><br> D.4
pochemuha

Answer:

A. 1

Explanation:

The higher you get in the atmosphere the cooler the temperature will be. The reason it's actually colder is because, as you go up in the atmosphere, the Earth's atmosphere feels less pressure the higher up you go. So as the gas in the atmosphere rises it feels less pressure, which makes it expand.

Hope this Help :)

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7 0
3 years ago
Rutherfordium-261 has a half-life of 1.08 min. How long will it take for a sample of rutherfordium to lose one-third of its nucl
Ira Lisetskai [31]

Answer:

t=1.712min

Explanation:

Hello!

In this case, since the radioactive decay equation is:

\frac{A}{A_0}=2^{-\frac{t}{t_{1/2} }

Whereas A stands for the remaining amount of this sample and A0 the initial one. In such a way, since the sample of rutherfordium is reduced to one-third of its nuclei, the following relationship is used:

A=\frac{1}{3} A_0

And we plug it in to get:

\frac{\frac{1}{3} A_0}{A_0}=2^{-\frac{t}{t_{1/2}} } \\\\\frac{1}{3}=2^{-\frac{t}{t_{1/2}} }

Now, as we know its half-life, we can compute the elapsed time for such loss:

log(\frac{1}{3})=log(2^{-\frac{t}{t_{1/2}} })\\\\log(\frac{1}{3})=-\frac{t}{t_{1/2}} }*log(2)

t=-\frac{log(\frac{1}{3})t_{1/2}}{log(2)} \\\\t=1.71min

Best regards!

8 0
2 years ago
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