Answer:
1.89 nol Cu(NO3)2
Explanation:
if you calculate it it will be 1.89
X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160
so, moles = 85.2 / (4y+160)
Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0
so, 48 = 16 x 10 x [85.2 / (4y+160) ]
Solve the equation to get y.
y = 31
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
<span>a regultate is to control or direct by a rule, principle, or method.</span>
