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Leya [2.2K]
2 years ago
5

A particle that has the electron configuration 1s²2s²2p³ would:

Chemistry
1 answer:
notka56 [123]2 years ago
5 0

Answer:

Have 2 filled orbitals and 3 partially filled orbitals.

Explanation:

Hello there!

In this case, according to the given information of the electron configuration for that particle; it is possible for us to infer it has 5 valence electrons, as the electrons on its outermost shell (2). Moreover, we undertand this particle needs three bonds, does not have neither the electron configuration of a noble gas which ends by p⁶ nor that of an alkali earth metal as it ends by s².

Therefore, we infer the correct answer is Have 2 filled orbitals and 3 partially filled orbitals because according to the Hund's rule, the s orbital is fulfilled and the p orbital has 1 electron orbital fulfilled and two partially filled orbitals.

Regards!

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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
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Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

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