<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
Answer:
alright bud lets see hmm.... the answer is a. 90.5kpa
Explanation:
92.3 kPa - 1.82 kPa = 90.5 kPa
keep up your hope also corrected me if a am wrong in anyway! :)
In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.
<h3>How is Sulphur hexafluoride formed?</h3>
Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.
Thus, a sample of fluorine yields 7g of SF6.
To learn more about Sulfur Hexafluoride click here;
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