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Romashka-Z-Leto [24]
4 years ago
5

The atoms of which element will NOT typically form ions?

Chemistry
2 answers:
tino4ka555 [31]4 years ago
7 0
He. it is a noble gas and therefore has a full outer shell of electrons. it does not need to gain or loose any.
ad-work [718]4 years ago
5 0

Answer: Option (C) is the correct answer.

Explanation:

Ions are the species that form when a neutral atom either gains or loses an electron.

For example, atomic number of beryllium is 4 and its electronic distribution is 2, 2. So, in order to attain stability by losing its valence electrons Be will form a Be^{2+} ion.

Whereas atomic number of helium is 2 and its electronic configuration is 1s^{2}.

Since, it has completely filled octet so, it will neither gain or lose an electron as it is stable in nature. Hence, it will not form ions.

Thus, we can conclude that the atoms He element will NOT typically form ions.

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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
Look at the diagram below.
damaskus [11]

Answer:

it will gain electrons to fill its outer shell

Explanation:

This element is boron which has 5 electrons.

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Answer:

Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.

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Answer:

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Explanation:

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