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andreyandreev [35.5K]
3 years ago
14

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Chemistry
1 answer:
Julli [10]3 years ago
7 0

Answer:

The number of molecules.

Explanation:

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what would you be most likely to measure by immsering and object in water and seeing how much the water level rises
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Mass, if you know what element you are working with.
4 0
3 years ago
Why is only one diastereomer formed in this reaction? Relate your answer to the mechanism you drew. b) If you used cis-stilbene
crimeas [40]

Answer:

1. Diastereomers have different physical properties (unlike most aspects of enantiomers) and often different chemical reactivity. ... Many conformational isomers are diastereomers as well. Diastereoselectivity is the preference for the formation of one or more than one diastereomer over the other in an organic reaction.

2. The result is a trans dibromide, as shown in the equation below

Explanation:

Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer.[1] Diasteoreomers are defined as non-mirror image non-identical stereoisomers. Hence, they occur when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.[2] When two diastereoisomers differ from each other at only one stereocenter they are epimers. Each stereocenter gives rise to two different configurations and thus typically increases the number of stereoisomers by a factor of two.

2. the addition of bromine to the trans and

cis isomers of 1,2-diphenylethene, more commonly known as trans- and cis-stilbene.

H

H

H H

trans-stilbene cis-stilbene

m.p. 122-124°C b.p. 82-84°C

density 0.970 g/mL density 1.011 g/mL

M.W. 180.25 g/mol M.W. 180.25 g/mol

In both cases, the nucleophilic double bond undergoes an electrophilic addition reaction

by the bromine reagent which proceeds via a cyclic bromonium ion. The addition of

bromine begins at one side of the double bond (either side is equally likely, but only one

option is drawn) and is followed by attack of bromide ion on the bromonium ion (again,

attack could occur at either carbon since the ion is symmetric, but only one option is

drawn). The result is a trans dibromide, as shown in the equation below:

Since the cis and trans isomers of stilbene have different geometries, it follows

that upon reaction with bromine they give rise to stereoisomeric bromonium ions and,

eventually, products that differ only by their stereochemistry.

4 0
3 years ago
1 point
inysia [295]

Answer:

The fungus has grown larger

Explanation:

Because where the orange is in the fridge and even normally you out oranges on the counter or in a bowl, where it's in the fridge it got old faster.

8 0
3 years ago
Question 21 of 25
RideAnS [48]

Answer:

7

Explanation:

7 0
3 years ago
Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c
LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass  of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H =  4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:  

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%  

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0
3 years ago
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