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Yanka [14]
1 year ago
9

A sample of a compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula of this

compound
Chemistry
1 answer:
Shtirlitz [24]1 year ago
5 0

Answer:

The empirical formula is CH2O, and the molecular formula is some multiple of this

Explanation:

In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C;  6.7⋅g1.00794⋅g⋅mol−1 H; and  53.5⋅g16.00⋅g⋅mol−1 O.

We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.

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A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de
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The density of the rectangular block in g/mL is 7.0.

<u>Given the following data:</u>

  • Mass of block = 22.8 gra1.94 kg
  • Length of block = 3.21 cm
  • Width of block = 5.83 cm
  • Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

Volume = length × width × height

<u>Conversion:</u>

1 in = 2.54 cm​

5.83 in = X cm

Cross-multiplying, we have:

X = 2.54(5.83)\\\\X = 14.81 \; cm

Volume = 3.21 × 5.83 × 14.81

Volume = 277.16 cubic centimeters.

<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}

<em>Density </em><em>= 7</em><em>.0 g/mL</em>

Therefore, the density of the rectangular block in g/mL is 7.0.

Read more: brainly.com/question/18320053

4 0
2 years ago
2
Alina [70]

Answer:

N - 1s²2s²2p³

Explanation:

Nitrogen is located in the p-block of the periodic table (groups 13-18) and is on the 2nd period.

The 2nd period tells us the principal energy level (a quantum number) is n = 2. Therefore, it must have already filled up the 1s sublevel.

The groups 13-18 on period 2 tells us that the 2s sublevel is also filled.

Nitrogen is located in Group 15. That means that there are 3 electrons that have filled the 2p sublevel, out of a possible 6.

Therefore, our electron configuration is 1s²2s²2p³

2p³ (Shorthand Config)

[He] 2s²2p³ (Noble Gas Config)

7 0
3 years ago
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