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3 years ago

Determine the empirical formula for a compound containing 85.6 g of carbon and 14.4 g of hydrogen. Please show work

Chemistry

1 answer:

photoshop1234 [79]3 years ago

4 0

Answer: $C_{1}H_{2}$

Explanation:

take gram & convert to mol (using the mm of the element) ,

85.6g C $\frac{1 mol }{12.01g}$ = 7.12 14.4g H $\frac{1 mol}{1.01g}$ =14.2 then take the smallest number & divide 7.12/7.12 =1 14.2/7.12= 2. since the numbers are in whole numbers the answer is $C_{1}H_{2}$ .

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PLeas Answer with A B C or D please be QUICK :>

Lena [83]

<em>Answer: </em>D

<em>Explanation:</em>

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electron configuration of C: 2,4

electron configuration of H: 1

there are 4 hydrogen atoms that donated 1 electron each to C

therefore it's D.

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2 years ago

If the pressure, volume, and the number of moles of a gas are known, which is needed to calculate the universal gas constant fro

umka21 [38]

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3 years ago

Which of the following electron configurations are written incorrectly?

Lynna [10]

Answer:

The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.

Explanation:

The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.

There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.

All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".

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3 0

3 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<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

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

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

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0

3 years ago

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oksano4ka [1.4K]

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6 0

3 years ago