Ionization energy is the energy required to remove the
outermost electron from one mole of gaseous atom to produce 1 mole of gaseous
in to produce a charge of 1. The greater the ionization energy, the greater is
the chance f the electron to be removed from the nucleus. In this casse, Radium
has the largest ionization energy.
1,3-pentadiene has two double bonds which are conjugated, which undergo electrophilic addition reaction on reacting with
.
The structure of 1,3-pentadiene is shown in the image.
When strong acid such as
reacts with 1,3-pentadiene, the electrophilic addition reaction can occur either on double bond at 1,2-position or at 3,4-position. The reaction that occurs is shown in the image.
Answer:
Zn+2HCl→ZnCl2+H2
Explanation:
Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.
NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)