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xxMikexx [17]
2 years ago
9

30 points!!

Chemistry
1 answer:
Lisa [10]2 years ago
7 0

Explanation:

hope this helps you

Mark my answer as brainalist

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What are two different kinds of rock that are easily weathered by carbonic acid?
horsena [70]
Limestone and marble are the two rocks that are easily weathered by carbonic acid. Two kinds of weathering mostly affect the rocks. and they are physical weathering and chemical weathering. The above mentioned two rocks are highly affected by carbonic acid. The appeareance of these kind of rocks change their structure due to erosion very regularly. Caronation is the type of chemical weathering that affects the rocks limestone and marble. The carbonic acide is formed by the reaction of carbon dioxide in the air and water in the rivers. This carbonic acid results in weathering.
8 0
3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135

8 0
3 years ago
You are given a solution of sugar and sand and water your task is to separate sugar from sand which of the following experimenta
Ostrovityanka [42]

Answer:

Explanation:

Filtration followed by evaporation:

To separate the mixture of sand and sugar, it is best to use the separation technique of filtration then evaporation.

Pour the water into the mixture. The sugar will dissolve with time in the water. Sand is made up of quartz and does not dissolve in water.

After the dissolution, filter the solution to separate the sand using a filter paper.

Dry the sand thereafter then proceed to evaporate the sugar with water solution. Evaporation will turn water into vapor and the sugar crystals will be left behind.

8 0
3 years ago
Which reaction model represents a nuclear fission reaction?
julsineya [31]

Answer: option B

Explanation: since nuclear fission involves the decay of larger nuclide into smaller nuclei along with Neutron when it is collide with Neutron.

Example Decay of U-235 into Kr and Ba along with 3 neutrons

8 0
3 years ago
Read 2 more answers
A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?
Tcecarenko [31]

Answer:

pH = 2.10

Explanation:

We name an acid as diprotic because it can release two protons:

H₂A  +  H₂O  ⇄  H₃O⁺   + HA⁻     Ka₁

HA⁻  +  H₂O  ⇄  H₃O⁺   + A⁻²      Ka₂

We propose the mass balance:

Analytical concentration = [H₂A] +  [HA⁻]  + [A⁻²]

As Ka₂ is so small, we avoid the [A⁻²] so:

0.18 M = [H₂A] +  [HA⁻]

But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

We propose the charge balance:

[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]

As we did not consider the A⁻², we can miss the term and if

Kw = H⁺ . OH⁻

We replace Kw/H⁺ = OH⁻. So the new equation is:

[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]

The acid is so concentrated, so we can avoid the term with the Kw, so:

[H₃O⁺] = [HA⁻]

In the mass balance we would have:

0.18 M = [H₂A]

We replace at Ka₁

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

Ka1 . 0.18 / [H₃O⁺] = [HA⁻]

We replace at the charge balance:

[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]

[H₃O⁺]² = 3.4×10⁻⁴  . 0.18

[H₃O⁺] = √(3.4×10⁻⁴  . 0.18)

[H₃O⁺] = 7.82×10⁻³

- log [H₃O⁺] = pH → - log 7.82×10⁻³

pH = 2.10

5 0
3 years ago
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