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OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.
Ratio 2:3
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
Zeff = Z - S
Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.
For boron, the electronic configuration is 1s₂ 2s₂ 2p₄
Z = 5, S = 2
Zeff = 5-2 = +3
For O, electronic configuration is 1s₂ 2s₂ 2p₄
Z = 8, S = 2
Zeff = 8-2 = +6
Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.
Answer:
The importance of crystal structure. The graphite-diamond mineral pair is an extreme example of the importance of crystal structure. These two very different minerals have exactly the same chemical formula (C), but the crystal structure of the two minerals is very different. In graphite, carbon atoms are bonded together along a flat plane, as shown in Figure 3.
