That would be an ionic bond since it has electrons to spare, rather than a convalent bond where all the electrons are taken up thus it is not a good conductor
Transient stage is also called the unstable stage and in this stage, there is a gradual decrease in deformation rate to a definite constant value.
<h3>What is Transient stage ?</h3>
A transient state is when a process variable or variables changes, but before the system reaches a steady state.
Transient time is the time it takes for a circuit to change from one steady state to the next.
Therefore, Transient stage is also called the unstable stage and in this stage, there is a gradual decrease in deformation rate to a definite constant value.
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ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
Answer:
Check the explanation
Explanation:
1)
Atomic number of Se = 34
So, number of proton = 34
use:
charge = number of proton - number of electron
-2 = 34 - number of electron
number of electron = 36
number of neutron = mass number - atomic number
= 78 - 34
= 44
Answer:
44
34
36
2)
number of proton = mass number - number of neutron
= 65 - 36
= 29
29 is atomic number for Cu
use:
charge = number of proton - number of electron
+1 = 29 - number of electron
number of electron = 28
Answer:
Cu
29
28
3)
Atomic number is 36 for Kr
use:
charge = number of proton - number of electron
= 36 - 36
= 0
use:
mass number = number of proton + number of neutron
= 36 + 42
= 78
Answer:
Kr
78
0
Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:
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