In which atmospheric layer is the ozone layer?
A.troposphere
B.mesosphere
C.stratosphere
D.thermosphere

Answers

Answer 1

Answer:

stratosphere

Explanation:

contains a high concentration of ozone in relation to other parts of the atmosphere, although still small in relation to other gases in the stratosphere.

Answer 2

Answer:

stratosphere

Explanation: Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface. Ozone is a molecule that contains three oxygen atoms.


Related Questions

The empirical formula of CBr2 has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound

Answers

Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain

Answers

Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

So, we're to first solve for H when [tex]C = 13.8[/tex]

Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

Convert to fraction

[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

Cross Multiply

[tex]3 * H = 13.8 * 8[/tex]

[tex]3 H = 110.4[/tex]

Solve for H

[tex]H = 110.4/3[/tex]

[tex]H = 36.8[/tex]

So, when

[tex]C = 13.8[/tex]

[tex]H = 36.8[/tex]

[tex]Total = C + H[/tex]

[tex]Total = 13.8 + 36.8[/tex]

[tex]Total = 50.6\ moles[/tex]

When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation

Answers

During sublimation

This has been posted on here before so you could’ve searched it lol.

Best of luck :))

Answer:

during sublimation

Explanation:

just took the test

Which is an example of a current research focus in chemistry?

A. applying gene therapy to treat certain diseases

B. using hook-and-loop tape in the clothing industry

C. developing smoke detectors for common use

D. studying coal combustion as an energy source

Answers

d if i’m not mistaken

Answer:

b is the correct answer

do not trust answer one

Explanation:

A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.

a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g

Answers

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

What is empirical formula?

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

To learn more about empirical formula, refer to the link below:

https://brainly.com/question/11588623

#SPJ2

   

How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The test variables (independent variables) and outcome variables (dependent variables) are the same things. B. The test variable (independent variable) controls the outcome variable (dependent variable). C. The test variable (independent variable) and outcome variable (dependent variable) have no affect on each other. D. The outcome variable (dependent variable) controls the test variable (independent variable).

Answers

Answer:

I'm on the exact same queston

Answer:

The test variable (independent variable) controls the outcome variable (dependent variable)

Explanation:

its right on study island

what’s the most abundant isotope of lawrencium

Answers

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

How to separate given mixture?

Answers

Answer:

Chromatography involves solvent separation on a solid medium.

Distillation takes advantage of differences in boiling points.

Evaporation removes a liquid from a solution to leave a solid material.

Filtration separates solids of different sizes.

Explanation:

If 25.6 mL isopropyl alcohol fully decomposes, what mass of H2 is formed? The density of isopropyl alcohol is 0.785 g/mL. g

Answers

Answer:

The correct answer is 0.67 g H₂

Explanation:

Isopropyl alcohol (C₃H₇OH) can decompose to give acetone (C₂H₆OH) and hydrogen gas (H₂) according to the following chemical equation:

C₃H₇OH (g) ⇒ C₂H₆CO(g) + H₂(g)

We can calculate the initial mass of isopropyl alcohol from the density and volume data:

density = m/V = 0.785 g/mL

⇒ m = density x V = 0.785 g/mL x 25.6 mL = 20.096 g C₃H₇OH

According to the chemical equation 1 mol of C₃H₇OH gives 1 mol H₂. The molar mass of C₃H₇OH is:

molar mass C₃H₇OH = (12 g/mol x 3) + (1 g/mol x 7) + 16 g/mol + 1 g/mol = 60 g/mol

molar mass H₂ = 1 g/mol x 2 = 2 g/mol

So, we obtain: 2 g H₂ from 60 g C₃H₇OH. We multiply this stoichiometric ratio (2 g H₂/60 g C₃H₇OH) by the initial mass of C₃H₇OH to obtain the mass of H₂ is formed:

20.096 g C₃H₇OH x (2 g H₂/60 g C₃H₇OH) = 0.6698 g ≅ 0.67 g H₂

The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .

Answers

Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

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