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Ray Of Light [21]
3 years ago
12

What are the systematic names for each of the structures alkanes

Chemistry
1 answer:
marta [7]3 years ago
7 0

Answer:

Name Molecular Formula Structural Formula

ethane C2H6 CH3CH3

propane C3H8 CH3CH2CH3

butane C4H10 CH3CH2CH2CH3

pentane C5H12 CH3(CH2)3CH

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The expected first intermediate formed during a halohydrin reaction is:
enot [183]

Answer:

B

Explanation:

The most stable carbonation with OH on the adjacent carbon

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3 years ago
What assumption is being made if scientists conclude that aspartic acid was formed by the prebiological synthesis in the passage
miss Akunina [59]

Answer:

A....

Explanation:

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3 years ago
Read 2 more answers
A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemic
Lostsunrise [7]
Phosphorus (5+), can have 5 bonds. It will have a double bond with Oxygen (2-) and single bonds with Chlorine (1-)

POCl3

* the 3 is a subscript
6 0
3 years ago
For an alloy that consists of 33.5 wt% Pb and 66.5 wt% Sn, what is the composition (a) of Pb (in at%), and (b) of Sn (in at%)? T
Tju [1.3M]

Answer:

Pb: 22.4 at%

Sn: 77.6 at%

Explanation:

It is possible to find at% of Pb and Sn converting mass in moles using molar mass assuming a basis of 100g, thus:

Pb: 33.5g × (1mol / 207.2g) = <em>0.1617mol</em>

Sn: 66.5g × (1mol / 118.7g) = <em>0.5602mol</em>

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Total moles: 0.1617mol + 0.5602mol = 0.7219mol

Composition in at%:

Pb: 0.1617mol / 0.7219mol × 100 = <em>22.4 at%</em>

Sn: 0.5602mol / 0.7219mol × 100 = <em>77.6 at%</em>

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I hope it helps!

5 0
3 years ago
3. In a typical titration experiment a student titrates a 5.00 mL sample of formic acid (HCOOH), a monoprotic organic acid, with
ira [324]

Answer:

2.893 x 10⁻³ mol NaOH

[HCOOH] = 0.5786 mol/L

Explanation:

The balanced reaction equation is:

HCOOH + NaOH ⇒ NaHCOO + H₂O

At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.

The moles of base added is calculated as follows:

n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH

Extra significant figures are kept to avoid round-off errors.

Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.

(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH

The concentration of HCOOH to the correct number of significant figures is then calculated as follows:

C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L

The question also asks to calculate the moles of base, so we convert millimoles to moles:

(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH

7 0
3 years ago
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