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Scilla [17]
3 years ago
6

Consider 80.0-g samples of two different compounds consisting of only carbon and oxygen. One of the compounds consists of 21.8 g

of carbon, and the other has 34.3 g of carbon. Determine the ratio in whole numbers of the masses of carbon that combine with 1.00 g of oxygen between the two compounds.
Chemistry
1 answer:
goldenfox [79]3 years ago
6 0

Answer:

  • <u><em>Ratio of the  mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2</em></u>

Explanation:

First, detemine the mass of oxygen in the two samples by difference:

  • mass of oxygen = mass of sample - mass of carbon

Item                     Compound 1                        Compound 2

Sample                80.0 g                                    80.0 g

Carbon                 21.8 g                                    34.3 g

Oxygen:               80.0 g - 21.8g = 58.2 g         80.0 g - 34.3 g = 45.7 g

Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:

  • For each sample, divide the mass of carbon by the mass of oxygen determined above:

Sample              Mass of carbon that combines with 1.00 g of oxygen            

Compound 1      21.8 g / 58.2 g =  0.375

Compound 2     34.3 g / 45.7 g = 0.751

Third, determine the ratio of the masses of carbon between the two compounds.

  • Divide the greater number by the smaller number:

  • Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2
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Answer:

See figure 1

Explanation:

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The problem states that the <u>terminal alkene</u> is the one that would is protonated. Therefore, we have to do the <u>protonation</u> in the double bond at the bottom to produce the <u>carbocation number 1</u>. Then, a hydride shift takes place to produce the <u>carbocation number 2</u>. A continuation, an <u>elimination reaction</u> takes place to produce the <u>conjugated diene</u>. Then the diene is protonated at the <u>carbonyl group</u> and with an elimination reaction of an hydrogen in the <u>alpha carbon</u> we can obtain <u>carvacol. </u>

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2 years ago
1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour
lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

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Metals are elements which are known by their special ability to form ions by a loss of electrons.

The increasing metallic character of metal is a measure of their ability to lose electrons.

Metallic character increases from right to left and down a group in the period table.

Metals are found to the left of the period table.

In conclusion, metals are known by their ability to lose electrons.

Learn more about metals at: brainly.com/question/25597694

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