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2 years ago

Identify the following as abiotic or biotic. A paramecium swimming in a lake

Chemistry

1 answer:

sertanlavr [38]2 years ago

4 0

<h3>Answer:</h3>

Biotic

<h3>Explanation:</h3>

We are being tested on whether we can differentiate between abiotic and biotic factors in an ecosystem.

We need to know that factors that affect living organisms in an ecosystem are classified into abiotic factors and biotic factors.

Then, what is the difference between abiotic and bioitic factors?

Abiotic factors are non-living factors that affect living organisms in an ecosystem while biotic factors are living factors that affect organisms in an ecosystem.

What are examples of Biotic and abiotic factors.

Examples of abiotic factors include wind, temperature, humidity, climate,. etc.
Examples of abiotic factors include parasitism, mutualism, commensalism, etc.

How do we arrive at the answer to our question?

A Paramecium swimming in a lake; this is a biotic factor.
Because paramecium is a living organism which relates with other organisms in the lake ecosystem in various way.

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3 years ago

Which of these is a TRUE statement? A) Sound travels through gases easier than liquids. B) Sound travels through solids easier t

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3 years ago

Read 2 more answers

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago

How many moles of lead, Pb, are in 1.50 x 10^12 atoms of lead?

gulaghasi [49]

<h3>Answer:</h3>

2.49 × 10⁻¹² moles Pb

<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>Atomic Structure</u>

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

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.50 × 10¹² atoms Pb

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.50 \cdot 10^{12} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})$
Multiply: $\displaystyle 2.49087 \cdot 10^{-12} \ moles \ Pb$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2.49087 × 10⁻¹² moles Pb ≈ 2.49 × 10⁻¹² moles Pb

5 0

3 years ago