The reaction between C2H6 and Cl2 is by addition

To calculate the molecular formula, convert the mass ratio into molar ratio as follows:

mass ratio of O:C=2:1

molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus, number of moles can be calculated as follows:

n=\frac{m}{M}

calculating ratio,

O:C=\frac{2 g}{16 g/mol}:\frac{1 g}{12 g/mol}=\frac{1}{8}:\frac{1}{12}=12:8=3:2

thus, molecular formula will be C_{2}O_{3}

Final answer:

If the ratio of oxygen to carbon by mass is 2.00:1.00, the formula of the oxide of carbon is CO₂ (carbon dioxide). This is because CO₂ has twice as much oxygen per amount of carbon as compared to CO (carbon monoxide). This situation adheres to the law of multiple proportions.

Explanation:

This question revolves around the concept of stoichiometry in chemistry, particularly in relation to the law of multiple proportions and finding the formula of an oxide of carbon given specific mass ratios.

Firstly, in carbon monoxide (CO), the ratio of oxygen to carbon by mass is 1.33:1.00.

However, when the ratio of oxygen to carbon by mass increases to 2.00:1.00, we are now dealing with a compound that contains twice as much oxygen per amount of carbon. In essence, this would be carbon dioxide (CO₂). The mass ratio of oxygen to carbon in CO₂ is indeed 2:1 (32 g/mol oxygen: 12 g/mol carbon).

This situation illustrates the law of multiple proportions - in this case, the two oxides of carbon (CO and CO₂) contain elements combined in ratios of small whole numbers.

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Final answer:

When 150 ml of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate, a silver chromate precipitate forms. Considering the stoichiometry of the reaction and the quantities of reactants, 24.88 grams of silver chromate will precipitate.

Explanation:

The subject of this question is based on precipitation reactions in Chemistry. Precipitation reactions occur when two solutions combine to form an insoluble solid known as a precipitate. The moles of silver nitrate present in a 150 mL of 0.500 M solution can be calculated using the formula Molarity = Moles ÷ Volume (in Litres).

Thus, Moles of AgNO3 = 0.500 M * 0.15 L = 0.075 mol AgNO3. According to the reaction equation 2AgNO3 + K2CrO4 → 2AgCrO4(precipitate) + 2KNO3, for every mole of K2CrO4, we have two moles of AgNO3. Thus, based on stoichiometry and the given quantities of the reactants, the limiting reactant will be AgNO3, and it will totally react and form the silver chromate precipitate. The moles of Ag2CrO4 formed would therefore also be 0.075 mol. To convert this into grams, we use the molar mass of Ag2CrO4, which is approximately 331.73 g/mol. Hence, grams of Ag2CrO4 = 0.075 mol Ag2CrO4 * 331.73 g/mol = 24.88 g Ag2CrO4.

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Answer:

A. Potassium (K)

Explanation:

The trend of atomic radius goes:

As group number increases, radius decreases

As period number increases, radius increases

Answer:

francium

Explanation:

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

Answer: Phosphorous

Explanation: Nitrogen(N): 7:

Phosphorus(P): 15:

Oxygen(O): 8:

Sulphur(S): 16:

As nitrogen contains half filled configuration, it is stable and does not react. Phosphorous lies below nitrogen and reactivity decreases down a group for non metals as they tend to gain electrons as the valence shell move away from the nucleus.

Oxygen has 4 valence electrons, thus it need to gain two electrons to attain stable configuration. Thus is reactive.

Sulphur lies below oxygen and also need to gain two electrons. Reactivity decreases down a group for non metals as they tend to gain electrons as the valence shell move away from the nucleus.