What is the scaling factor if the molar mass of
1 answer:
Answer:
The scaling factor is 5.
Explanation:
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In this case, since the scaling factor is defined as the ratio of the molar mass of the molecular formula (complete) to the empirical formula (simplified), it is possible to compute it for the empirical formula of CH2O whose molar mass is 30 g/mol (12+2+16) as shown below:
Therefore, we can also infer that the molecular formula would be:
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Answer:
The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.
Explanation:
To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.
So you know:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
So, the molar mass of NaOH is:
NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole
Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?
mass= 11.2 grams
<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.</em></u>
(a): This is an elimination reaction. The reaction proceeds in 3 steps; Step 1: Protonation of alcohol, step 2: Loss of leaving group, and step 3: Deprotonation to form an alkene. Elimination (E1) reaction dominates than substitution (SN1) reaction because the formed HSO4- anion is a poor nucleophile and tends not to add to the carbocation intermediate formed. Hence the product formed is I, is trans.
(b) This is a dihydroxylation of the alkenes reaction. A catalytic amount of OsO4 is used along with an oxidizing agent, which oxidizes the reduced osmium(VI) into osmium(VIII) to regenerate the catalyst. The reaction proceeds in 2 steps; Step 1: Cis addition of OsO4 to double bond and form 4 member cyclic osmate ester. Step 2: Hydrolysis of the intermediate to 1.2-diol. Hence the product formed is II, is cis.
(c) This is hydroboration - oxidation reaction. The reaction proceeds through the addition of H- on the more substituted carbon and BH2 on the less substituted carbon of the double bond. The reaction is a syn addition. Hence the product formed is III.
(d) This a reduction reaction where carboxylic acid (-COOH) is reduced to 1o alcohol (-CH2OH). Hence the formed product is IV.
Stereochemistry, a subdiscipline of chemistry, involves the take a look at the relative spatial association of atoms that form the shape of molecules and their manipulation.
The observation of stereochemistry specializes in stereoisomers, which by means of definition have an equal molecular method and series of bonded atoms (constitution), however, differ within the three-dimensional orientations of their atoms in space. for that reason, it's also referred to as 3-d chemistry three-dimensionality.
Learn more about Stereochemistry here:-brainly.com/question/13266152
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<u>Disclaimer:- your question is incomplete, please see below for the complete question.</u>
Each of the following reactions has been reported in the chemical literature. Predict the product in each case, showing stereochemistry where appropriate.
