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vladimir1956 [14]
3 years ago
10

Smog usually forms from gound-level ozone and what other human-made pollutant?

Chemistry
2 answers:
Bingel [31]3 years ago
4 0

This happens when pollutants emitted by cars, power plants, industrial boilers, refineries, chemical plants, and other sources chemically react in the presence of sunlight. Ozone at ground level is a harmful air pollutant, because of its effects on people and the environment, and it is the main ingredient in “smog."

arsen [322]3 years ago
3 0

The word smog is created from the words "smoke" and "fog". It contains chemicals such carbon monoxide, sulfur dioxide, and various nitrogen oxides. It is created mostly from the burning of fossil fuels, such as gasoline, which is used commonly as fuel components for many different vehicles. Having known that such chemicals are found in exhaust fumes from cars, it is safe to say that the answer could either be smoke, or carbon dioxide(depending on teacher expectations).

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What is the process called when a solid turns to a gas
maxonik [38]
When a solid turns to gas it is called sublimation, and when a gas turns into a liquid it is called deposition 
5 0
3 years ago
Read 2 more answers
Jada Peterson
barxatty [35]

Answer: A

Explanation:

4 0
2 years ago
What titrant will be used to titrate the 0. 02 m hcl phenol red solution?
RideAnS [48]

The NaOH will be used What titrant  to titrate the 0. 02 m hcl phenol red solution.

Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.

Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.

Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.

To know more about titrant

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7 0
1 year ago
Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th
soldi70 [24.7K]

Answer: from the Zn anode to the Cu cathode


Justification:


1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)


2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).


3) Then, you can already tell that electrons go from Zn to Cu.


4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.


So you get that the electrons flow from the anode (Zn) to the cathode (Cu).


Always oxidation occurs at the anode, and reduction occurs at the cathode.

3 0
3 years ago
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
2 years ago
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