Answer: The structure of
1-Butyne is shown below and it contains;
Nine Sigma Bonds Two Pi Bonds
Explanation:
In given structure the sigma bonds are highlighted with blue colour and pi bonds are highlighted by Black colour and indicated by red arrows.
Remember the very first covalent bond between two atoms is always the sigma bond. After formation of sigma bond if further p orbitals with unpaired electrons are available they will form pi bond, or two pi bonds as in given scenario.
Answer:
0 M.
Explanation:
Hello,
In this case, the undergoing reaction is:
![M(NO_3)_2+NaCN\leftrightarrow [M(CN)_4]^{-2}+NaNO_3](https://tex.z-dn.net/?f=M%28NO_3%29_2%2BNaCN%5Cleftrightarrow%20%5BM%28CN%29_4%5D%5E%7B-2%7D%2BNaNO_3)
Nonetheless, it only matters the reaction forming the given complex:
![M^{+2}+4CN^-\leftrightarrow [M(CN)_4]^{-2}](https://tex.z-dn.net/?f=M%5E%7B%2B2%7D%2B4CN%5E-%5Cleftrightarrow%20%5BM%28CN%29_4%5D%5E%7B-2%7D)
In such a way, the formation constant turns out:
![K_F=\frac{[[M(CN)_4]^{-2}]_{eq}}{[M^{+2}]_{eq}[CN^{-}]_{eq}^4}](https://tex.z-dn.net/?f=K_F%3D%5Cfrac%7B%5B%5BM%28CN%29_4%5D%5E%7B-2%7D%5D_%7Beq%7D%7D%7B%5BM%5E%7B%2B2%7D%5D_%7Beq%7D%5BCN%5E%7B-%7D%5D_%7Beq%7D%5E4%7D)
Now, one could assume that the initial concentrations of the ions equals the original compounds concentrations:
![[M^{+2}]_0=0.150M;[CN^-]_0=0.820M](https://tex.z-dn.net/?f=%5BM%5E%7B%2B2%7D%5D_0%3D0.150M%3B%5BCN%5E-%5D_0%3D0.820M)
In such a way, we modify the formation constant in terms of the change 
due to the reaction progress:
Now, solving for :
The feasible solution is 0.15M which will lead to an equilibrium concentration of M⁺² of 0M
![[M^{+2}]_{eq}=0.15M-0.15M=0M](https://tex.z-dn.net/?f=%5BM%5E%7B%2B2%7D%5D_%7Beq%7D%3D0.15M-0.15M%3D0M)
This fact has sense since the formation constant is very large.
Best regards.
Answer: 1:The bowling ball have more gravitational potential energy as it sit on top of the building
Explanation:
Because the exponent in the 10s term is positive move the decimal point 4 places to the right:
1.2 ×104 =12,000
The question is incomplete, the complete question is;
Consider the following reaction at 298K.
2 H+(aq) + 2 Cr2+(aq) =H2(g) + 2 Cr3+(aq)
Which of the following statements are correct?
Choose all that apply.
K > 1
ΔGo < 0
Eocell < 0
n = 2 mol electrons
The reaction is reactant-favored.
Answer:
K > 1
ΔGo < 0
n = 2 mol electrons
Explanation:
If we look at the reaction; 2 H+(aq) + 2 Cr2+(aq) -----> H2(g) + 2 Cr3+(aq), we will notice that the reaction is a spontaneous reaction because E°cell= 0-(-0.41) = 0.41 V
If the reaction is a spontaneous electrochemical process, we expect it to be product favoured with K>1. Also, a spontaneous reaction must have ∆G<0.
Lastly, we can see that two electrons were transferred according to the balanced equation for the electrochemical reaction.
