Use the periodic table to answer the questions below.

Chemistry

1 answer:

madreJ [45]2 years ago

5 0

The diagram that show the correct electron configuration is the second option. ⬆⬇1s, ⬆⬇2s, ⬆⬆⬆2p.

<h3>What is electronic configuration?</h3>

The electronic configuration is the representation of electron present indifferent energy levels is an atom. The different type of orbitals are s, p, d, and f.

Thus, the correct option is 2, ⬆⬇1s, ⬆⬇2s, ⬆⬆⬆2p.

Learn more about electronic configuration

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What is a substance?

shusha [124]

I think the answer is C. An element or compound that cannot be physically separated. Sorry if im wrong.

7 0

4 years ago

If the vapor pressure of pure benzene is 96.1 mm Hg, what must the vapor pressure of pure toluene be in order for the 50/50 % mi

mylen [45]

Answer:

$P_{tol}=30.34mmHg$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the vapor pressure of pure toluene by using the Raoult's law as shown below:

$y_{tol}P_{mix}=x_{tol}P_{tol}\\\\y_{ben}P_{mix}=x_{ben}P_{ben}$

Thus, we solve for the mole fraction of benzene in the vapor phase first:

$y_{ben}=\frac{x_{ben}P_{ben}}{P_{mix}} =\frac{0.5*96.1mmHg}{63.2mmHg}=0.76$

Which means that the mole fraction of toluene in the vapor phase is 0.24, and therefore, the vapor pressure of pure toluene turns out to be:

$P_{tol}=\frac{y_{tol}P_{mix}}{x_{tol}} =\frac{0.24*63.2mmHg}{0.5}=30.34mmHg$

Regards!

7 0

3 years ago

Consider the decomposition of the compound C5H6O3 as follows: C5H6O3(g)  C2H6(g) + 3CO(g) A 5.63 g sample of pure C5H6O3(g) was

deff fn [24]

Answer:

$K_{eq}=1.02x10^{-4}$

Explanation:

Hello,

In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:

$5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3$

After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:

$n_{total}^{eq}=\frac{P_{total}^{eq}V}{RT}=\frac{1.63atm*2.50L}{0.082\frac{atm*L}{mol*K}*473.15K} =0.105mol$

Then, by knowing the moles at the equilibrium considering the change " $x$ ", which yields to:

$\ \ \ \ \ C_5H_6O_3(g) \leftrightarrow C_2H_6(g) + 3CO(g)\\I\ \ \ \ \ 0.0494mol\ \ \ \ \ \ 0mol \ \ \ \ \ \ \ \ 0mol\\C\ \ \ \ \ \ -x\ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x\\E\ \ 0.0494mol-x\ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ 3x$