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2 years ago

What is ozone hole? What causes an ozone hole?

Chemistry

2 answers:

lutik1710 [3]2 years ago

6 0

Answer:

Ozon prevents us from UV rays of the sun. Ozone hole means that UV rays are entering the Earth. This results in cancer.

Explanation:

Chlorofluorocarbons are causing this

Pie2 years ago

4 0

Answer:

The term ‘ozone hole’ refers to the depletion of the protective ozone layer in the upper atmosphere (stratosphere) over Earth's polar regions. People, plants, and animals living under the ozone hole are harmed by the solar radiation now reaching the Earth's surface—where it causes health problems, from eye damage to skin cancer.

Ozone depletion occurs when chlorofluorocarbons (CFCs) and halons—gases formerly found in aerosol spray cans and refrigerants—are released into the atmosphere.

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Q7-A graduated cylinder is filled to the 12.0 mL line with water. A solid with a mass of 14.52 g is placed in the graduated cyli

Tems11 [23]

Answer:

If it served you, give me 5 stars please, thank you!

6 0

2 years ago

Describe the relationship between the radius of a cation and that of the atom from which forms.

vichka [17]

The radius of the cation is much smaller than the corresponding neutral atom.(b) The radius of an anion is much larger than the corresponding neutral atom.Explanation:The size of the atom or ion is inversely proportional to the nuclear charge experienced by the electrons.(a)The size of the cation is smaller than the size of the corresponding neutral atom. This is because after removal of an electron from the highest principle energy level the nuclear charge experienced by the valence electrons increases resulting in the decrease in size.(b)The size of an anion is larger than the size of the corresponding neutral atom. In an anion, an extra electron is added to the highest principle energy level but the effective nuclear charge pulling the electrons towards the nucleus is still same. The net effective nuclear charge experienced by the electrons present in the outermost shell decrease. Moreover, due to the added electron, the repulsion between the electrons also increases resulting in the increase in size

Make since? i hope this helps

4 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?

Luba_88 [7]

The empirical formula is XeO₃.

Explanation:

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/ 131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0

3 years ago

I need help with question 48

Stells [14]

Answer:

V H2SO4=30 ml=0.03 l

Cm=16M

CM=n/V

n=CM×V

n= 0.48 moles H2SO4

Cm=0.48/0.3

CM=1.6M

Answer B

6 0

2 years ago