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Feliz [49]
3 years ago
12

An analysis of a sugar compound is found to contain 132 grams of carbon (C), 22 grams of hydrogen (H), and 176 grams of oxygen (

O). Determine the empirical formula of the compound.
Chemistry
1 answer:
ddd [48]3 years ago
8 0
132 g of C  ,   22 g of H   , 176 g of O

132 + 22 + 176 => 330 g <span>of the substance

</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u        H = 1.0 u       O = 16.0 u

C = 132 / 12.0 => 11 moles

H = 22 / 1.0 => 22 moles

O = 176 / 16.0 => 11 moles

Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1

H = 22 / 11 => 2

O = 11 / 11 => 1

 formula empirically <span>is : CH</span>₂O

hope this helps!


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Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Selec
meriva

Answer:

Central atoms with four electron groups will be sp3 hybridized. True

Hybrid orbitals are delocalized over the entire molecule. False

The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True

Atoms with a single pi bond and an octet are sp2 hybridized. True

Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False

All resonance structures must be considered when assigning hybridization. False

Explanation:

When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.

Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.

According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.

When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.

Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.

The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.

8 0
3 years ago
(b) in what way are carbon dioxide and orange juice similar?​
ArbitrLikvidat [17]

Answer:

The effects of supercritical CO2 (SC-CO2) on the microbiological, sensory (taste, odour, and colour), nutritional (vitamin C content), and physical (cloud, total acidity, pH, and °Brix) qualities of orange juice were studied. The CO2 treatment was performed in a 1 litre capacity double-walled reactor equipped with a magnetic stirring system. Freshly extracted orange juice was treated with supercritical CO2, pasteurised at 90°C, or left untreated. There were no significant differences in the sensory attributes and physical qualities between the CO2 treated juice and freshly extracted juice. The CO2 treated juice retained 88% of its vitamin C, while the pasteurised juice was notably different from the fresh juice and preserved only 57% of its vitamin C content. After 8 weeks of storage at 4°C, there was no microbial growth in the CO2 treated juice.

6 0
2 years ago
What is the mass number for an element that has 62 protons, 85 neutrons, and 62 electrons
Nata [24]

Answer:

The elements mass number is 147

Explanation:

5 0
3 years ago
Use the periodic table to identify the number of core electrons and the number of valence electrons in each case below.
Harman [31]

Can you please add a picture of the case??

5 0
2 years ago
The half life of carbon is 5,730 years. What fraction of the original 14/6 C would be in a wooden axe handle that was 17,190 yea
Mariana [72]

Answer:

\frac{1}{8}  

Explanation:

The half-life of carbon (5730 y) is the time it takes for half the carbon to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

\text{17 190 y} \times \frac {\text {1 half-life}}{\text{5730 y}} = \text{3 half-lives}  

We can construct a table.

<u>No of half-lives</u>  <u>Fraction remaining</u>

            1                       \frac{1}{2}  

            2                      \frac{1}{4}  

            3                      \frac{1}{8}  

The general formula is

\text{Fraction remaining} = \frac{1}{2^{n}}

where <em>n</em> = the number of half-lives.

Thus, \frac{1}{8} of the original carbon remains after 17 190 y.

8 0
3 years ago
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