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Helen [10]
2 years ago
14

What compound is this?

Chemistry
1 answer:
Nezavi [6.7K]2 years ago
7 0

Answer:

Carbon Tetrachloride

Explanation:

1 Carbon atom, 4 chlorine atoms (hence "tetra" prefix)

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What are the advantages of having proof before you make any conclusion about your own understanding for scientific method?
Kobotan [32]

Answer:

Helps in accepting and defending the conclusion.

Explanation:

The main advantages of having proof before you make any conclusion is that it gives a reason for accepting the answer given by you because you have the proof to prove your answer so no one questions on your answer. The proof give you strength to defend your conclusion and make your conclusion right in front of the world. Without proof no one can accept your answer because every answer of a scientific question wants a proof for its validation.

3 0
3 years ago
A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
Cloud [144]

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

=-345272 j

= -345.2 KJ

5 0
2 years ago
A certain automobile contains 4 tires, 2 headlights, 1 steering wheel,
Step2247 [10]

Answer:

143 parts I think probably not

5 0
2 years ago
When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final react
velikii [3]

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

  • Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.  
  • Propagation; free radicals react with molecules to produce new free radicals and molecules.
  • Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}

\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet

\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}

\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}

\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}

\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet

Termination

\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}

8 0
3 years ago
Please help me, I really don't want to fail but I don't know how to do this
Brrunno [24]

Answer:

A)

<u>4, 7, 4, 6</u>

B)

<u>12 moles</u>

Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

3 0
2 years ago
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