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3 years ago

HELP!!!!!!!!!!!!!...!.!.!!..!.!

Chemistry

2 answers:

RSB [31]3 years ago

6 0

Answer : The correct option is, (C) 2-octene

Explanation :

Structural formula or bond-line formula : In the structural formula, the bonding and type of bonds which holds the atoms in molecule together are shown.

In bond-line formula, the lines are used between the bonded atoms and the atoms are also shown.

The basic rules for naming of organic compounds are :

First select the longest possible carbon chain.

For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.

A suffix '-ane' is added at the end of the name of alkane.

If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

In the given hydrocarbon, the longest possible carbon chain number is 8 that means we add prefix 'oct' and suffix '-ene' for double bond. So, the name of given molecule will be, 2-octene.

Hence, the name of the given molecule is, 2-octene.

just olya [345]3 years ago

4 0

Answer:

option b is the answer

Explanation:

so the of the molecules

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What is required in order to determine whether or not an object moves?

AysviL [449]

Answer:

A reference point.

Explanation:

Reference point:

A reference point is a point which is used to determine weather the object is in motion or not. The moving object is compared with the reference point.

An object is in the state of motion when its distance is changed relative to the other object or reference point.

In order to determine weather an object is moving or not a point of reference is needed.

Without reference point displacement can not be measured

3 0

3 years ago

Read 2 more answers

Would your results be he same if 1.5 g of barium chloride was used instead of 1.3 g? same with 0.7 g of barium chloride?

pickupchik [31]

For a chemical reaction, the results would be the same with different amounts of barium chloride present if barium chloride is not the limiting reactant of the reaction. The limiting reactant is the substance that is totally consumed when the reaction proceeds to completion. In other words, it dictates the amount of products that will be produced. Otherwise, if barium chloride is the limiting reactant, the amount of barium chloride present would definitely affect the results.

6 0

3 years ago

Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

2 years ago

A4 kg object is being pulled to the east by a 6 N force. What is the object's acceleration?

Dahasolnce [82]

1.5 ms⁻²

Explanation:

We understand that Force is also given as mass * acceleration;

F = Ma

If force is 6N and the mass is 4kg of the object, the a can be evaluated as follows;

6 = 4a

6/4 = a

1.5 = a

= 1.5m/s²

4 0

3 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the ratio of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago