What type of consumer are bears?

1 answer:

5 0

Bears are omnivores (and scavengers), meaning they eat both plants and animals.
it can be both a primary consumer (if it eats plants) and secondary & tertiary consumers (if it eats a plant-eating animal). so it depends on what it eats.

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Air has a mass of 1.2g and a volume of 4,555ml. What is the density

coldgirl [10]

Answer:

<h2>Density = 0.00026 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3> $Density(\rho) = \frac{mass}{volume}$ </h3>

From the question

mass of air = 1.2 g

volume = 4,555 mL

Substitute the values into the above formula and solve for the density

That's

<h3> $Density = \frac{1.2}{4555}$ </h3>

= 0.0002634

We have the final answer as

<h3>Density = 0.00026 g/mL</h3>

Hope this helps you

3 0

3 years ago

CH4 + O2 → CO2 + H2O

Scilla [17]

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

<u>Atomic Structure</u>

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

<u>Organic</u>

Naming carbons

<u>Stoichiometry</u>

Analyzing reaction rxn
Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

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

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] Molar Mass of H: 1.01 g/mol

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O$