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Mars2501 [29]
3 years ago
14

Imagine a molecule with six total domains that are confined to two-dimensional space, so the electron group arrangement is hexag

onal planar. If a molecule contained two lone domains and 4 bonded domains, in what positions would the lone domains be located: 1 and 2, 1 and 3, or 1 and 4

Chemistry
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

extra bonds don't count

Explanation:

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Ca (ClO) 2<br> Spell out the full name of the compound.
antoniya [11.8K]
Calcium hypochlorite
5 0
3 years ago
Determine the chemical formula for the molecule shown.
OLEGan [10]

Answer: Formula is C4H8O. In every branch of line formula there are Carbon atom. Carbon makes 4 bonds so In branch left in which two CH3-groups are attached there are also an Hydrogen aton which is not shown.

Explanation: Molecule is organic, it is classified as aldehyde.

Hydrocarbons contains only Carbon and Hydrogen atoms, Organic molecules can contain also other atoms.

Reactions of aldehydes are reduction to alcoholes or oxidation to carboxylic acids. It can not for a polymer.

5 0
2 years ago
Write the balanced standard combustion reaction for the c5h7on3s2
AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0
2 years ago
Which step in the free-radical chlorination of methane do you expect to be the most exothermic? attempt this problem without loo
stiks02 [169]

The termination step of the free-radical chlorination of methane is the most stable one among all three steps.

The free-radical substitution reaction between chlorine and methane features three major steps:

Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus <em>endothermic</em>.

Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g., \cdot \text{CH}_3. The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.

Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.

Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.

3 0
3 years ago
Water is placed in a graduated cylinder and the volume is recorded as 43.5 ml. a homogeneous sample of metal pellets with a mass
natulia [17]
Density = mass /volume of the body. mass = 10.88 g, volume of the liquid displaced= 49.4-43.5=5.9 ml.

density = 10.88/5.9=1.844 g/ml~1.84 g/ml

The volume displaced is exactly the same as that of the body (The Eureka fro Archimedes!)
7 0
2 years ago
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