<u>The troposphere: </u>
H. This layer can have thunderstorms or clear, sunny skies.
A. The biosphere interacts most with this layer.
<u>The stratosphere:</u>
B. It is the second layer from Earth's surface.
G. Winds are strong and steady in this layer.
<u>The mesosphere:</u>
E. It is heated by the ozone layer beneath it.
D. This layer is where most meteor showers occur.
<u>The thermosphere :</u>
F. It contains the ionosphere and exosphere.
C. It contains layers of single, unmixed gas.
<u>Explanation:</u>
Depending on the Earth's temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.
Stratosphere (10-30) includes much of the surface ozone. The change in height temperature arises as this ozone absorbs ultraviolet (UV) radiation from the sun. The temperature in Mesosphere (30-50 Km) declines again with height, hitting a minimum of about -90 ° C at the "mesopause." Above this thermosphere (50-400 Km) is settled which is a area where temperatures rise with height once again. The penetration of intense UV and X-ray radiation from the sun induces this temperature rise.
Answer:
The volume of the stock solution needed is 1L
Explanation:
Step 1:
Data obtained from the question. This include the following:
Concentration of stock solution (C1) = 6M
Volume of stock solution needed (V1) =?
Concentration of diluted solution (C2) = 1M
Volume of diluted solution (V2) = 6L
Step 2:
Determination of the volume of the stock solution needed.
With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
6 x V1 = 1 x 6
Divide both side by 6
V1 = 6/6
V1 = 1L
Therefore, the volume of the stock solution needed is 1L
Answer:
16L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 4L
P1 (initial pressure) = 4atm
P2 (final pressure) = 1atm
V2 (final volume) =?
Using Boyle's law equation P1V1 = P2V2, the new volume can be obtain as follow:
P1V1 = P2V2
4 x 4 = 1 x V2
16 = V2
V2 = 16L
Therefore, her new lungs volume is 16L
