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3 years ago

What is the source of energy used to react chlorine with methane?

Chemistry

2 answers:

Papessa [141]3 years ago

8 0

Ultraviolet light is the correct answer

Luden [163]3 years ago

5 0

Answer:

The source of energy used to react with chlorine and methane is ultraviolet rays or in the presence of sunlight.

Explanation:

The source of energy used to react with chlorine and methane is ultraviolet rays or in the presence of sunlight.This reaction is an example of Photo-chemical reaction.

Photo-chemical reaction is the type of chemical reaction in which energy is imparted to the reactants by the light or reactions initiated by light.

The photo chemical reaction of methane with chlorine in presence of ultraviolet light is given as:

$CH_4+Cl_2\overset{h\nu}\rightarrow CH_3Cl+HCl$

$CH_3Cl+Cl_2\overset{h\nu}\rightarrow CH_2Cl_2+HCl$

$CH_2Cl_2+Cl_2\overset{h\nu}\rightarrow CH_1Cl_3+HCl$

$CH_1Cl_3+Cl_2\overset{h\nu}\rightarrow CCl_4+HCl$

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Identify the right

Basile [38]

.answer:

so the first one is analogous

the second is homologous

and the third is analogous

explanation:

analogous. this means they share a similar function (flight) but do not have the same embryonic origin.

6 0

3 years ago

Pls help

Irina-Kira [14]

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0

3 0

2 years ago

Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma

Lina20 [59]

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

$-CH_2=CH-CH=CH2-$

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

8 0

3 years ago

A student observes chloroplasts in the micrograph of a cell.

allochka39001 [22]

Answer:

I think it's fungus or plant cell

6 0

2 years ago

Read 2 more answers

What volume of 0.350 m koh is required to react completely with 24.0 ml of 0.650 m h3po4?

BartSMP [9]

The complete balanced chemical equation for this is:

First we calculate the number of moles of H3PO4:

moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol

From stoichiometry, 3 moles of KOH is required for every mole of H3PO4, therefore:

moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole H3PO4) = 0.0468 mol

Calculating for volume given molarity of 0.350 M KOH:

Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7 mL

Answer:

7 0

3 years ago