Answer: <span>A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called
<em>cis</em> Isomer.
Explanation: Geometric isomerism takes place about the double bond in alkenes when the alkyl groups are either situated at the same side (<em>
cis</em>) or are situated opposite (
<em>trans</em>) to each other.
Example: <em>
cis</em>
-2-Butene (highlighted red)
<em>
trans</em>
-2-Butene (highlighted blue)</span>
Explanation: cell in the body is enclosed by a cell (Plasma) membrane. The cell membrane separates the material outside the cell, extracellular, from the material inside the cell, intracellular. ... All materials within a cell must have access to the cell membrane (the cell's boundary) for the needed exchange
**Answer**: The answer would be Yes I believe
There are a few things you can do to make slime less sticky. You can either:
1) add a bit of baking soda (one to three tea spoons depending on how big it is, for the one that you show I would say one should be enough)
2)You can place it in hot water and squish the water out of it before placing it in cold water and then drying it out a bit with a towel. Don't forget to knead it a bit so it stays smooth and not too hard.
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
