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Akimi4 [234]
2 years ago
7

Ozone located in the troposphere is _______. A. Considered a pollutant b. Beneficial to living organisms c. Higher in concentrat

ion than ozone in the stratosphere d. All of the above Please select the best answer from the choices provided A B C D.
Chemistry
1 answer:
Aleonysh [2.5K]2 years ago
3 0

The presence of ozone in the troposphere has been considered as pollutant. Thus, option A is correct.

The atmosphere has been divided into five layers based on the availability and the composition of different elements.

<h3>Troposphere</h3>

The troposphere has been the lowest layer of the atmosphere. It has been closed to the earth surface and provides with clouds and maintain weather.

The gases on the troposphere are used by the organisms for living purpose. The ozone has been a harmful gas for living organisms.

Thus, the presence of ozone in the troposphere has been considered as pollutant. Thus, option A is correct.

Learn more about troposphere, here:

brainly.com/question/201658

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of
Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0
2 years ago
aluminum has a density or 2.7g/cm3. copper has a density of 8.96/cm3. which metal would you choose to build a model airplane?
erik [133]

Answer: Aluminium

Explanation: Aluminium metal has a lower density than copper. So, for the same volume of metal used to build a model airplane, the aluminium plane would be very lightweight while that of copper would be heavy.  The lightweight airplane will fly easily.

7 0
2 years ago
Easy Questions! Chemistry - Please help! I give thanks
Evgen [1.6K]
This is not chemistry but

see it is a triangular prim on it's side
V=BH
are of base times height
the base is a triangle
height is 18.5

base=1/2bh
b=8.6
h=8.4
base=1/2(8.6)(8.4)
base=36.12

V=bh
V=36.12*18.5
V=668.22

round
V=668.2 ft^3
8 0
2 years ago
Read 2 more answers
Density (D) is defined as the ratio of mass (m) to volume (V) and can be determined from the expression D = . Find the density o
olchik [2.2K]

Explanation:

87.329 g/32.32 cm³ = 2. 70 g/cm³

4 0
2 years ago
How many lead (Pb) atoms are in a 5.32 g sample of pure lead?
Elanso [62]

Answer:

1.55×10²² molecules.

Explanation:

We'll begin by calculating the number of mole in 5.32 g of pure lead (Pb). This can be obtained as follow:

Mass of Pb = 5.32 g

Molar mass of Pb = 207 g/mol

Mole of Pb =?

Mole = mass /molar mass

Mole of Pb = 5.32/207

Mole of Pb = 0.0257 mole

Finally, we shall determine the number of molecules in 0.0257 mole of Pb. This can be obtained as follow:

From Avogadro's hypothesis,

I mole of Pb contains 6.02×10²³ molecules.

Therefore, 0.0257 mole will contain = 0.0257 × 6.02×10²³ = 1.55×10²² molecules.

Therefore, 5.32 g of pure lead (Pb) contains 1.55×10²² molecules.

4 0
3 years ago
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