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3 years ago

Which of the following is NOT an example of chemical weathering?

Chemistry

2 answers:

ololo11 [35]3 years ago

8 0

Answer: d. ice wedging in rock

Explanation:

Weathering is a natural process in which the rocks are broken down into pieces and sediments due to the effect of physical and chemical agents. Chemical weathering is a process by which the rocks are subjected to the treatment of chemical reaction like oxidation and hydrolysis which makes the physical structure of the rocks weak and causes the disintegration of rocks.

d. ice wedging in rock is the correct option this is because of the fact that it is the example of physical weathering instead of chemical weathering. In the case of ice wedging in rocks the water inside the rocks get solidified inside the rocks which leads to weakening of internal structure of rocks. Hence, results in weathering.

lara [203]3 years ago

6 0

The answer is D. an ice wedging in Rock
Mainly because that is a Mechanical reaction or Physical

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The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

4 years ago

Mark as brainlist asap plz i need the answer asap

Darya [45]

A
Please mark as brainliest
Emily

5 0

4 years ago

Determine whether each carbohydrate is best described as a monosaccharide, a disaccharide, or a polysaccharide

vazorg [7]

Answer options from an alternative source

fructose
lactose
starch
glucose
cellulose

Answer:

fructose -monosaccharide
lactose - disaccharide
starch - polysaccharide
glucose - monosaccharide
cellulose - polysaccharide

Explanation:

Monosaccharides are carbohydrates that are the simplest form of a sugar. They cannot be further broken down into smaller carbohydrates, and represent the basic building block for carbohydrates. Monosaccharides can form disaccharides, which are the sugar formed when two monosaccharides join together, or polysaccharides, which are chains of monosaccharides.

3 0

3 years ago

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

5 0

3 years ago

What are the benefits and drawbacks of using algae as a fuel source?

I am Lyosha [343]

Answer:The Benefit is that it is a good natural fuel for nature and it will not hurt the environment

Explanation:

The drawback is that it will not last long and I don’t know which fuel you mean either for tools and cars,etc or for food but for cars,etc like o said it will not last long and for food it can be poison to the human body

8 0

3 years ago

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