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Gre4nikov [31]
3 years ago
13

Which is an example of chemical weathering?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

b. the layer of oxide formed on statues made of metal

Explanation:

Chemical weathering is a process whereby rocks are disintegrated or weathered causing a chemical reaction thereby leading to change in the parent rock constituents .

In chemical weathering a new substance is usually formed after the weathering takes place.

From the question the layer of statue made by metal will likely form oxides of those metals. Example Silver will react with oxygen to form an entirely new substance like silver(i)oxide.

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Do Alkali Metal Compounds form precipitates? And Why?
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Answer:

Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.

6 0
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Numerator and denominator are Equal to one another? <br> I don’t get the concept
Alekssandra [29.7K]

It's like saying that if 1 meter is equal to 100 cm,

\frac{100cm}{1m}

or

\frac{1m}{100cm}

You can base the answers from the chart

5 0
3 years ago
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An example of a solid aerosol
Ulleksa [173]
Examples are smoke, oceanic haze, air pollution, smog, etc.
7 0
3 years ago
Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.
Bezzdna [24]

Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

            NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

8 0
3 years ago
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