Answer:
See Explanation
Explanation:
10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.
11) The bombardment of U-235 with neutrons leads to the reaction;
Hence
a = 92, b= 95, c= 53
12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.
Hence;
Answer: option D) energy was absorbed and entropy increased.
Explanation:
1) Given balanced equation:
2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).
2) Being the energy placed on the side of the reactants means that the energy is used (consumed or absorbed). This is an endothermic reaction.
So, the first part is that energy was absorbed.
3) As for the entropy, it is a measure of the disorder or radomness of the system.
Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.
With that, you have shown that the right option is D) energy was absorbed and increased.
Answer:
A
Explanation:
If its going at a constant speed it will not accelarate wich means to speed up.
Answer:
D) 
Explanation:
Carbon.
The electronic configuration is -
Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.
But in methane, 
it forms 4 bonds. So, 1 electron each from 2s orbital jumps to the next orbital in the p subshell.
Thus, the configuration is:-
Thus, the valence electron configuration is:-
<u>Answer:</u> The concentration of 
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of 
, we use the equation:
Moles of = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
For the given chemical equations:
![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:
The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As, is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of required will be 0.285 M.
