Consider the balanced equation K,Croa (aq) + Pb(NO3)2 (aq) → PbCrO4(s) + 2KNO3(aq)

Identify the number of each type of atom in this formula 5C12H22O11<br><br>wdym

Lady bird [3.3K]

There are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in one molecule of C12H22O11.
In five molecules there are 60 carbon atoms, 110 hydrogen atoms, and 55 oxygen atoms.

6 0

2 years ago

A sample of C3H8 has 6.72 x 10^24 H atoms. <br> What is the total mass of the sample?

gregori [183]

The answer is 492.8 g

1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample.

1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole

6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n

n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol

2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:

Ar(C) = 12 g/mol

Ar(H) = 1 g/mol

Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol

3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:

m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g

6 0

2 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

2 years ago

If we use 18 moles of H2 how many moles of H2O do we make? Use this chemical equation: 6H2 + O2 → 3H2O

Ilya [14]

<h3>Answer:</h3>

9 mol H₂O

<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>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6H₂ + O₂ → 3H₂O

[Given] 18 mol H₂

[Solve] mol H₂O

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

[RxN] 6 mol H₂ → 3 mol H₂O

<u>Step 3: Stoich</u>

[DA] Set up conversion: $\displaystyle 18 \ mol \ H_2(\frac{3 \ mol \ H_2O}{6 \ mol \ H_2})$
[DA] Simplify: $\displaystyle 18 \ mol \ H_2(\frac{1 \ mol \ H_2O}{2 \ mol \ H_2})$
[DA] Divide [Cancel out units]: $\displaystyle 9 \ mol \ H_2O$