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2 years ago

1) How many aluminum atoms are there in 3.50 grams of Al2O3?

Chemistry

1 answer:

drek231 [11]2 years ago

3 0

Answer:

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms = moles aluminium * Number of Avogadro

Aluminium atoms = 0.0686 * 6.022 * 10^23

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

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1. john needs to create a buffered solution at a ph of 3.5 for his biomedical laboratory

Lunna [17]

Answer:

Use a ratio of 0.44 mol lactate to 1 mol of lactic acid

Explanation:

John could prepare a lactate buffer.

He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.

$\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}$

He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.

For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.

6 0

1 year ago

CCl, 189.86 g/mol. Express your answer as a chemical formula.

kompoz [17]

Answer is: a chemical formula of the compound is C₄Cl₄.

CCl is empirical formula of the compound. Empirical formula gives the proportions of the elements present in a compound, so we know that proportion od carbon and chlorine is 1 : 1 (n(C) : n(Cl) = 1 : 1).

M(CCl) = Ar(C) + Ar(Cl) · g/mol.

M(CCl) = 12.0107 + 35.453 · g/mol.

M(CCl) = 47.4637 g/mol; molar mass of empirical formula.

M(compound) = 189.86 g/mol; molar mass of the compound.

M(compound) / M(CCl) = 189.86 g/mol / 47.4637 g/mol.

M(compound) / M(CCl) = 4; chemical formula of the compound have four carbon and four chlorine atoms.

5 0

1 year ago

When a 375 mL sample of nitrogen is kept at constant temperature, it has a pressure of 1.2 atmospheres. What pressure does it ex

kolezko [41]

Answer:

It exerts a pressure of 3.6 atm

Explanation:

This is a gas law problem. We are looking at volume and pressure with temperature being kept constant, thus, the gas law to use is Boyle’s law. It states that at a given constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically; P1V1 = P2V2

Let’s identify the parameters according to the question.

P1 = 1.2 atm

V1 = 375mL

P2 = ?

v2 = 125mL

We arrange the equation to make room for P2 and this can be written as:

P2 = P1V1/V2

P2 = (1.2 * 375)/125

P2 = 3.6 atm

3 0

1 year ago