The atmosphere of earth is made of five main layers.
1) Troposphere : This is the lowest part of the atmosphere. Most of the air that makes up the atmosphere is present in this layer.
2) Stratosphere : This layer is present above troposphere and extends up to 50 km. It contains ozone layer which prevents the harmful ultraviolet rays from the sun from entering the lower troposphere
3) Mesosphere : The layer above stratosphere is known as mesosphere.
4) Thermosphere : The region lies above mesosphere.
5) Exosphere : The is the outermost region of the atmosphere.
From the above discussion we can see that the layer that lies between exosphere and mesosphere is Thermosphere
70.33 L is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.
<h3>What is volume?</h3>
Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.
Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.
We will use ideal gas law to find the volume
PV = nRT
Can also be written as
V = (nRT)/P
Where,
P = pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature
Here, we have given
P = 3.5 atm
V = to find
n = 10 moles
R = 0.08206 L⋅atm/K⋅mol
T = 300k
Lets substitute the values
V = (10 × 0.08206 × 300)/3.5
V = 70.33 L
Learn more about volume
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Answer:
D. Hydrogen bonds between complementary base pairs89
Explanation:
A DNA molecule is composed of two long polynucleotide chains made of four types of nucleotide subunits, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). These nucleotides are joined by covalent bonds forming a phosphate-sugar backbone. <em>These strands are held to one another with hydrogen bonds between the base portions of complementary nucleotides.</em>
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The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
