An example.
water is H2O
2 hydrogen, 1 oxygen
so the number to the right means how much of what is on the left.
so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)
so 6 carbons, 12 hydrogen
the ratio of c2 to h4 doesn't change it's always 1:2.
but the 3 at the front is a different number relating to how much you have
The 1st law of thermodynamics doesn't specify that matter can be created nor destroyed, but that the total amount of energy in a closed system cannot be created nor destroyed though it can be changed from one form to another.
The correct matches are:
1. Exosphere - Temperatures reach as high as 2000 C yet it feels very cold
This is the top layer of the atmosphere. The atoms are so dispersed that despite it having very high temperature it doesn't feel like it at all.
2. Thermosphere - Particles that have enough energy can escape into space
The thermosphere is the fourth highest layer of the atmosphere. The atoms in this layer are relatively distant from one another, so the particles that have enough energy manage to escape easily into the exosphere and then the space.
3. Mesosphere - It is the coldest region of the atmosphere
The mesosphere is the third highest layer. In this layer the temperatures constantly drop, and they go down to -85 degrees, making it the coldest layer by far.
4. Stratosphere - Ninety percent ozone is in this layer
The startosphere has a separte zone in it which is dominated by only one gas, the ozone. It is called the ozone layer, the one that protects the Earth from too intense UV radiation, and in fact over 90% of this gas is locate here.
5. Troposphere - It is warm due to the heat from Earth's surface
The troposphere is the densest and lowest of the layers. It is the one that also has Greenhouse gases which manage to trap the heat that is radiated from the surface of the Earth, thus keeping this layer relatively warm.
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
For the first blank, that is the endoplasmic reticulum
For the second, it is lysosome
For the third blank, it is the cell membrane
For the fourth, sorry I don’t know this one
For the fifth, that is the vacuole
For the sixth, that is mitochondrion
For the seventh, that is Golgi body
And lastly the eighth, it is the nucleus
Sorry I did not know what the fourth was but everything else is good.
