Answer:
The 3p orbitals have the same general shape and are larger than 2p orbitals, but they differ in the number of nodes. You have probably noticed that the total number of nodes in an orbital is equal to n−1 , where n is the principal quantum number. Thus, a 2p orbital has 1 node, and a 3p orbital has 2 nodes.
Answer:
<h3>When a drought occurs, their food supply can shrink and their habitat can be damaged. ... Losses or destruction of fish and wildlife habitat. Lack of food and drinking water for wild animals. Increase in disease in wild animals, because of reduced food and water supplies.</h3><h3>While insects and cacti might provide a meagre supply of water, most desert animals survive by being what Price calls "water misers". ... To perform this feat, they have evolved specialized kidneys with extra microscopic tubules for extracting water from urine.</h3>
Explanation:
The wave length of light. This determinds the color/type of light.
Answer is: the discovery of sub atomic particles like electrons.
J. J. Thomson discovered the electron in 1897.
His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.
With this model, he abandoned his earlier hypothesis (the atom was composed of immaterial vortices).
J.J. Thomson placed two oppositely charged electric plates around the cathode ray. He did experiments using different metals as electrode materials and found that the properties of the cathode ray remained constant no matter what cathode material he used.
Tomson concluded that atoms are divisible and that the corpuscles are their building blocks (atoms are made up of smaller particles).
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
