Answer:
The reaction is endothermic.
Yes, absorbed
3.06x10¹kJ are absorbed
Explanation:
In the reaction:
2HgO(s) → 2Hg(l) + O₂(g) ΔH = 182kJ
As ΔH >0,
<em>The reaction is endothermic</em>
<em />
As the reaction is endothermic, when the reaction occurs,
<em>the heat is absorbed.</em>
<em></em>
Now, based on the equation, when 2 moles of HgO (Molar mass: 216.59g/mol), 182kJ are absorbed.
72.8g are:
72.8g * (1mol / 216.59g) = 0.3361 moles HgO.
that absorb:
0.3361 moles HgO * (182kJ / 2 moles) =
<h3>3.06x10¹kJ are absorbed</h3>
ANSWER
A standard light microscope is used to view living organisms with little contrast to distinguish them from the background, which would be harder to see with the electron microscope.
Electron microscopes can be used to examine not just whole cells, but also the subcellular structures and compartments within them.
Epsom salt is MgSO4.
We assume x water of hydration in the crystalline form.
Molecular mass of MgSO4 = 24+32+4*16=120
Molecular mass of MgSO4.xH2O = 120+18x
By proportion,
2.000/0.977 = (120+18x)/120
Cross multiply
0.977(120+18x) = 120*2.000
from which we solve for x
17.586x+117.24 = 240
x=122.78/17.586
=6.980
Answer: there are 7 water of hydration in Epsom salt, according to the experiment.
Note: more accurate (proper) results may be obtained by using exact values (3-4 significant figures) in the molecular masses. However, since water of hydration is the nearest integer, using approximate values (to at least two significant figures) suffice.
Answer:
hey mateee
Boyle's law :- the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature.
Boyle's law can also be formularized as P1V1 = P2V2
Charle's law :- the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.
Charle's law can also be formularized as V1/T1 = V2/T2
Answer:
During complete oxidation of the fatty acid CH3(CH2)14COOH, eight molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven times.
Explanation:
In the β-oxidation of fatty acids, an acetyl-CoA molecule is removed from the fatty acid chain after every β-oxidation cycle that the fatty acid undergo, leaving behind a fatty acyl-CoA molecule shortened by two cabon atoms..
The removal of the acetyl-CoA molecule starts from the carboxyl end and shortens the fatty acid molecule by two carbon units. Successive β-oxidation cycles results in the complete oxidaton of the fatty acid molecle to acetyl -CoA molecules.
The compound CH3(CH2)14COOH, is a 16-carbon saturated fatty acid molecule known as palmitic acid.. It undergoes seven passes through the β-oxidation cycle to yield eight molecules of acetyl-CoA with each cycle yielding an acetyl-CoA molecule and a fatty acyl-CoA shortened by two carbon atoms. Finally, the seventh step yields two acetyl-CoA molecules.
