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3 years ago

A particle that has the electron configuration 1s²2s²2p³ would:

Chemistry

1 answer:

notka56 [123]3 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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If 200 g of water is heated from 20 C to 40 C, then what is the heat added in kilocalories (same as food calories)?

Maslowich

Answer: 4Kcal

Explanation:

H= mcø

M=200g

C= 1 cal/g/°c

Ø= 40-20=20°c

H= 200*1*20= 4000calories= 4Kcal

8 0

3 years ago

How many molecules are in 85g of silver nitrate?

maksim [4K]

<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Writing Compounds
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})$
Multiply/Divide: $\displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

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3 years ago

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3 years ago

Read 2 more answers

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