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inessss [21]
3 years ago
5

Identify 6 examples of weathering (mechanical, chemical) in your environment such as (household, yard, street, cemetery, park, a

nd playground). Identify what type of weathering occurred. (Ex. Oxidation, dissolution/carbonation). Explain how each example occurred. Usea picture to show each example of the processe
Chemistry
2 answers:
Bogdan [553]3 years ago
4 0
Chemical weathering is when things get weathered chemically 
kirill115 [55]3 years ago
3 0
Chemical, water in metal, wood and water, aging and tree
mechanical, wind on house wind on tree dust and walls
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Which statement is always true about a reversible chemical reaction?
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Answer: b

Explanation:

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You try to measure out exactly 5.0 milliliters of water by eye into each of five test tubes. When you go back to check the volum
jeka94

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The volumes are both, accurate and precise.

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3 years ago
There are four major methods of purification in the chemistry laboratory. Which of the given techniques is not one of the four m
RideAnS [48]
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Sublimation

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6 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
2 years ago
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