14. Calculate the number of (a) gram atoms in 4.6 gram of sodium.
(b) Oxygen atoms in 0.10 mole of sodium carbonate decahydrate
Answers
Answer:
Explanation:
1.(a)
Since 1 mole of Na2CO3.10H20 = 6.022 X1023molecules of Na2CO3.10H20
So 0.1 mole of Na2CO3.10H2O will have = 6.022 X1023 X0.1 molecules of Na2CO3.10H20
Now we know that each molecule contains 13 O-atoms,
Total no. of O- atoms = 6.022 X1022 X 13
hence O2 molecules will be = 6.022 X1022 X 13 /2= 6.022 X1022 X 6.2
Now further, 6.022 X1023 molecules of Oxygen = 1 mole Oxygen gas .
Thus, 6.022 X1022 X 6.2 molecules of Oxygen will contain= 6.022 X1022 X 6.2 / 6.022 X1023 moles of oxygen
= 0.62 mole of oxygen
Now as one mole oxygen = 32 gm of oxygen
So, 0.62 mole oxygen = 32 X 0.62= 19.84 gm of Oxygen gas.
(b)
6.022*1023atoms weight=32g
1 atoms weight=32/6.022*1023
(c)(NH4)2SO4
2..
cathode rate were composed of a previously unknown negatively charged particle, which was later named the electron .Cathode ray tubes(CRTs) use a focused beam of electrons deflected by electric or magnetic fields to create the image on a television screen it was started by a J.J Thomson.
4.Properties of cathode rays:-
They are deflected from their rectilinear path by electrostatic and magnetic fields and behave in the manner of a stream of negatively charged particles.
They can impart negative charge to objects in their paths.
They can cause ionization in gases.
They cast shadows if some target is placed in their path.
The negatively charged particles are electrons. The mass and charge of an electron is same irrespective of the substance (element or compound) it is generated. Hence, e/m or charged mass ratio is constant and independent of the type of gas.
3.(a)
•An electron can make a transition from a stationary state of higher energy E2 to a state of lower energy E1and in doing so, it emits a single photon of frequency,
ν=E2−E1h where h is the constant.
Conversely, on absorbing an energy, E2−E1 when the electron is at energy E1, the electron can make a transition from E1 to E2.
•Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h2π,
Thus, L=nh2π where n=0,1,2,..
Answer:
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Explanation:
