Help me out please an thank you

Chemistry

1 answer:

ivanzaharov [21]3 years ago

3 0

The second one, I could be mistaken though

You might be interested in

Can someone show me how to do this?

Pachacha [2.7K]

To start off any questions, we write down the givens.

Mass CO2 = 168.2g
Molar Mass CO2 = <span>44.011g/mol
Moles CO2 = ?

To find the number of moles, we do mass/molar mass to find the # moles

n CO2 = 44.011g / 168.2 g/mol
n CO2 = </span>0.26165874 mol
n CO2 = 0.2617 mol

The number of moles in the sample is 0.2617 mol.

5 0

3 years ago

jolli1 [7]

Answer:

B. first level consumers.

Explanation:

The first level consumers are also the primary consumers. They are the animals that depends or rely on plants to transform chemical energy into carbohydrates and sugar.

Primary consumers are mostly herbivores. They feed on plant matter by which they derive their nutrition. They have special enzymes that helps to convert plant materials into substances needed for their body.

So, the first level consumer takes plant matter and convert them into useful food materials for their own nutrition.

The second level consumers depend on the first level consumers for nutrition.

3 0

3 years ago

How many molecules are contained in 103.4g of sulfuric acid?

Ierofanga [76]

<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Compounds</u>

Writing Compounds
Acids/Bases

<u>Atomic Structure</u>

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

<h3>Explanation:</h3>

<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

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

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 103.4 \ g \ H_2S(\frac{1 \ mol \ H_2S}{34.09 \ g \ H_2S})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2S}{1 \ mol \ H_2S})$
Multiply: $\displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S$