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3 years ago

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Chemistry

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

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Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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