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3 years ago

In which layer of the atmosphere is the ozone layer?

2 answers:

8 0

Ozone is mainly found in two regions of the Earth's atmosphere. Most ozone (about 90%) resides in a layer that begins between 6 and 10 miles (10 and 17 kilometers) above the Earth's surface and extends up to about 30 miles (50 kilometers). This region of the atmosphere is called thestratosphere.

jarptica [38.1K]3 years ago

7 0

Hello,

The correct answer is D) Troposphere

The troposphere is the first layer above the surface and contains 70 to 80 per cent of the Earth's atmosphere. Weather occurs in this layer. Many jet air crafts fly in the stratosphere because it is very stable. Also, the ozone layer absorbs harmful rays from the Sun.

Hope this helps!!!! :)

**(Vanessa)**

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

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3 years ago

Help me outttttttttttt t t t. T t

Alla [95]

Answer:

the open bag allowed the gas from the reaction to escape

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3 years ago

A compound has the empirical formula CH2O. If the compound’s molecular mass is 180 g/mol, determine the molecular formula of the

Umnica [9.8K]

The molecular formula will be a multiple of the empirical CH2O. One unit of CH2O has a mass of 12+2*1+16 = 30 g. This means that if our compound has a molecular mass of 180 g/mol, we can divide 180 / 30 = 6 units, and our compound has 6 units of CH2O. This means that its molecular formula is C6H12O6.

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3 years ago

Gases and particles which are put into the air or emitted by various sources are called __________. photochemical smog emissions

Leto [7]

I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.

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3 years ago

What is the electron configuration of the calcium ion?

Elanso [62]

[Ar] 4s²
Let me know if you want a step by step!
Hope that helps

4 0

3 years ago

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