11.2L/22.4L (STP value) x 1 mol of CH4 x 16.04 g of CH4 = 8.2 g
Answer:
V₂ = 15.6 L
Explanation:
Given data:
Initial volume = 175 mL (0.175 L)
Initial pressure = 1 atm
Initial temperature = 273 K
Final temperature = -5°C (-5+273 = 268 K)
Final volume = ?
Final pressure = 1.16 kpa (1.16/101=0.011 atm)
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm
V₂ = 46.9 L / 3.003
V₂ = 15.6 L
Answer:
Amino acids, along with glucose, are reabsorbed in the glomerular system with a passive or active mechanism as the fluid travels through the entire renal tubular system and enters the circulation again.
Active mechanisms are those that require expenditure of energy, that is, expenditure of the energy currency, while the passive ones do not, they occur through spontaneous non-energy processes such as osmosis, the osmotic gradient and the difference in concentrations in different compartments.
Explanation:
Glomerular filtration is the regulator of the excretion of metabolites and toxic molecules or not necessary for our body. That is why if the amino acid values are high as well as those of glucose in urine, we will be facing a pathology.
If glucose is increased, it is because there is a glycemic peak in blood volume, hence possible diabetes.
And if the amino acids are increased, we could be facing an autoimmune or proteolytic pathology where a large amount of body proteins such as muscle proteins would be breaking down and releasing the amino acids that make it up, this phenomenon usually appears in those people who suffer from rhabdomyolysis in expenses very intense energy sources not appropriate.
On the other hand, glomerular filtration occurs in the kidney and is carried out by the nephron, which is the functional unit of the kidney, within it there is a specific tubular system in collection, absorption and reabsorption, added to the presence of Bowman's capsule.
Problem One (left)
This is just a straight mc deltaT question
<em><u>Givens</u></em>
m = 535 grams
c = 0.486 J/gm
tf = 50
ti = 1230
Formula
E = m * c * (ti - tf)
Solution
E = 535 * 0.486 * ( 1230 - 50)
E = 535 * 0.486 * (1180)
E = 301077
Answer: A
Problem Two
This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.
E = H m
H = 2257 J/gram
m = 11.2 grams
E = 2257 * 11.2
E = 25278 to three digits is 25300 Joules. Anyway it is the last one.
Three
D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.
Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.
Answer: B
Four
The best way to think about this question is just to get the answer. Answer C.
A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.
B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.
C: is correct because the molecules neither stop nor do they stick. The hit and move on.
D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.
E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.
Five
Half life number 1 would leave 0.5 grams behind.
Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.
Answer: 0.25
Answer C
Check the attached file for the answer.