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2 years ago

What is the volume of 35.0 g of propylene glycol if the density of propylene glycol is 1.036 g/ cm3?

1 answer:

3 0

Mass = 35.0 g

density = 1.036 g/cm³

volume = ?

therefore:

D = m / V

1.036 = 35.0 / V

V = 35.0 / 1.036

V = 33.784 cm³

answer D

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Water can be formed from the synthesis reaction of hydrogen with oxygen<br><br> PLEASE HELP

Anestetic [448]

Answer:

1.4 mols

4th answer

Explanation:

22. 5 g of O2 in moles = (22.5/32) mols = 0.703 mol

The stoichiometry between O2 and H2O =1: 2

Therefore H2O produced = 2 * 0.703 mols=1.406 mols

8 0

2 years ago

For a principal quantum number n, how many atomic orbitals are possible?

slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

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2 years ago

How many molecules are in 85g of silver nitrate?

maksim [4K]

<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Writing Compounds
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})$
Multiply/Divide: $\displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0

2 years ago

A molecule contains 24.36 g of nitrogen and 62.64g of sliver

Komok [63]

Answer:

This molecule is AgN3

Explanation:

5 0

2 years ago

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A chemist determined by measurements that 0.050 moles of aluminum participated in a chemical reaction. Calculate the mass of alu

Law Incorporation [45]

Answer:

13.5 g

Explanation:

This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.

Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.

number of moles = n = mass of Al / Atomic Weight Al

⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹

= 13.5 g

We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.

8 0

3 years ago