Answer:
Follows are the solution to this question:
Explanation:
Please find the image file of the chemical reaction in the attachment:
In a water medium, the CH3- type CH 3Li is a heavy nucleophile that attacks the carbonyl carbon atom to form the alkoxide ion, which will then be protonated to form alcohol.
We have been given the condition that carbon makes up 35%
of the mass of the substance and the rest is made up of oxygen. With this, it
can be concluded that 65% of the substance is made up of oxygen. If we let x be
the mass of oxygen in the substance, the operation that would best represent
the scenario is,
<span> x = (0.65)(5.5 g)</span>
<span> <em> </em><span><em>x =
3.575 g</em></span></span>
Sulfur-32 Information
This element is absorbed by plants from the soil as sulphate ion. Sulfur has 23 isotopes, 4 of them are stable. Sulfur is used in matches, gunpowder, medicines, rubber and pesticides, dyes and insecticides.
<h3>Which isotope of sulfur contributes the least to its mass number?</h3>
Sulfur (16S) has 23 known isotopes with mass numbers ranging from 27 to 49, four of which are stable: 32S (95.02%), 33S (0.75%), 34S (4.21%), and 36S (0.02%).
<h3>What is the use of Sulphur 35 isotope?</h3>
A radioactive sulfur isotope; a beta emitter with a half-life of 87.2 days; used as a tracer in the study of the metabolism of cysteine, cystine, methionine, and other compounds; also used to estimate, with labeled sulfate, extracellular fluid volumes.
Learn more about isotopes here:
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brainly.com/question/364529</h3><h3 /><h3>#SPJ4</h3>
I think the correct answer would be the third option. The reason I2 has a higher melting point than F2 is because I2 possesses a more polarizable electron cloud. I2 contains more electrons than F2 which would result to a stronger intermolecular forces. Having stronger intermoleculer forces would mean more energy is needed to break the bonds so a higher melting point would be observed.
The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
