CHEMISTRY COLLEGE

Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equation.

Answers

Answer 1

Answer:

Answer:

30 cm³

Explanation:

Step 1: Given data

Density of aluminum (ρ): 2.7 g/cm³

Mass of aluminum (m): 81 g

Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

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Previous

Answers

Answer:

Abe chutiye, Question sahi se likh laude phle

A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.

Answers

Answer:pH = 2.96

Explanation:

C5H5N + HBr --------------> C5H5N+ + Br-

millimoles of pyridine = 80 x 0.3184 =25.472mM

25.472 millimoles of HBr must be added to reach equivalence point.

25.472 = V x 0.5397

V =25.472/0.5397= 47.197 mL HBr

total volume = 80 + 47.197= 127.196 mL

Concentration of [C5H5N+] = no of moles / volume=

25.472/ 127.196= 0.20M

so,

pOH = 1/2 [pKw + pKa + log C]

pKb = 8.77

pOH = 1/2 [14 + 8.77 + log 0.20]

pOH = 11.0355

pH = 14 - 11.0355

pH = 2.96

5.0 liters of a gas are at an initial pressure of 5.0 atmospheres. If the temperature and amount of a gas are kept constant, what is the new volume of the gas when pressure is increased to 7.0 atmospheres?

Answers

Final answer:

Using Boyle's Law of gases which states that the pressure and volume of a gas have an inverse relationship when temperature is kept constant, we find that when the pressure of the gas increases from 5.0 to 7.0 atmospheres, the volume of the gas decreases to approximately 3.57 liters.

Explanation:

The question pertains to the application of Boyle's Law, a fundamental concept in the field of physics dealing with gases. Boyle's Law states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. This means if the pressure of a gas increases, the volume decreases, and vice versa.

In this case, you have 5.0 liters of a gas under an initial pressure of 5.0 atmospheres. The pressure is then increased to 7.0 atmospheres, and you are asked to determine the new volume of the gas. To solve this problem, we use the formula for Boyle's Law, which is P1V1 = P2V2. We know P1 (initial pressure) is 5.0 atmospheres and V1 (initial volume) is 5.0 liters. P2 (final pressure) is increased to 7.0 atmospheres and V2 (final volume) is what we are trying to find.

So, we plug the numbers into the equation and get: 5.0 atmospheres * 5.0 liters = 7.0 atmospheres * V2. Solving for V2, we find V2 to be approximately 3.57 liters. Therefore, when the pressure of the gas is increased from 5.0 atmospheres to 7.0 atmospheres, the volume decreases to around 3.57 liters, while the temperature remains constant.

Learn more about Boyle's Law here:

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How to atoms share more than one pair of electrons?

Answers

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

More than one pair of electrons can be shared between atoms to form double or triple covalent bonds. Unlike ionic bonds, covalent bonds are often formed between atoms where one of the atoms cannot easily attain a noble gas electron shell configuration through the loss or gain of one or two electrons.

Salt, sugar, and snow are examples of ____________________ solids.

Answers

Crystalline I hope this helps:)

They are examples of Semi solids

The vapor pressure of carbon disulfide is 355.6 torr at 25°C. What is the vapor pressure of a solution prepared by dissolving 10.60 g naphthalene (C10H8, Molar Mass = 128.2 g/mol) in 155 mL CS2 liquid (Molar Mass = 76.14 g/mol, density = 1.261 g/mL)? Assume the solution obeys Raoult's law, and treat naphthalene as a nonvolatile solute.