According to Boyle's law, what would happen to the pressure of a gas if the temperature were tripled as the number of moles and the volume were held constant? A. The pressure would be one-third of its original value.
B. The pressure would be one-ninth of its original value.
C. The pressure would triple.
D. The pressure would remain the same.

Answers

Answer 1

Answer:

According to Boyle's law, if the temperature were tripled as the number of moles and the volume were held constant, the pressure would triple (option C).

What is Boyle's law?

Boyle's law is the observation that the pressure of an ideal gas is inversely proportional to its volume at constant temperature.

However, when the temperature of a gas is increased, the pressure of the gas also increases provided the volume is constant.

According to this question, the temperature of a gas tripled as the number of moles and the volume were held constant.

Therefore, according to Boyle's law, if the temperature were tripled as the number of moles and the volume were held constant, the pressure would triple.

Learn more about Boyle's law at: brainly.com/question/1437490

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Why do olive oil and vinegar (a water-based solution) tend to separate in salad dressing? will added salt dissolve in the oil or in the vinegar? explain your answer?

Answers

Vinegar (acetic acid solution) is a polar substance, whereas olive oil is non polar in nature. Like dissolves like. A polar substance cannot dissolve in a non polar compound. Therefore, vinegar which is polar is not soluble in olive oil, a non polar substance. Hence, in a salad dressing olive oil separates out from vinegar.

Added salt would dissolve in vinegar solution as salt, a polar substance which is chemically, sodium chloride is an ionic substance. It can dissolve in the polar vinegar solution but not olive oil.

HCl solution, about 250 mL in volume. The pH of the solution is 1.37. Calculate the molar concentration of HCl in this solution. Please use the scientific presentation of the numerical answers (i.e., 1.23x 102), and round your answers to two decimal places.

Answers

Answer:

Molar concentration of HCl in this Solution = 0.01 molar

Explanation:

Given pH = 1.37

So Concentration of H⁺ is equal to $10^(- 1.37)$ = 0.042 molar

Volume of HCl solution = 250 ml

Since,

1000 ml solution contain 0.042 mole of HCl

250 ml solution contain = $(0.042 X 250)/(1000)$ = 0.01 molar

Round off value = 0.01

Molar concentration of HCl in this Solution = 0.01 molar

Compare the reactivity of alcohols with the length of the R-group.

Answers

The longer the alkyl chain is, the lesser the reactivity of the alcohol and vice versa.

What are alcohols?

Alcohols are substances that possess the -OH functional groups. We know that the reactivity of the alcohol depends on the length of the alkyl group.

Now, the longer the alkyl chain is, the lesser the reactivity of the alcohol and vice versa. This is why the rate of reaction of high molecular weight alcohols is low.

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As you increase the chain of a alcohol, the reactivity decreases. This is due to an increase in disorder and decrease in intermolecular attraction between molecules.

All chemical reactions use reactants in a specific proportion or stoichiometry to form products. The __________ reactant regulates the amount of products produced. A) excess B) limiting C) proportional D) stoichiometric

Answers

The correct answer is B) Limiting.

Hope this helped!

Answer:

B: Limiting

Explanation:

The limiting reactant regulates the amount of products produced, because you have the least amount of the limiting reactant.

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms?

Answers

Answer:

C. It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation:

took test

It may have twice as much electrons a the potassium ion.

Consider the reaction C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/∘C. The temperature increase inside the calorimeter was found to be 22.0 ∘C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose.