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Tju [1.3M]
4 years ago
12

How does a ketone react with PCl5?,

Chemistry
1 answer:
Reptile [31]4 years ago
7 0

When ketone is reacted with phosphorous pentachloride, chlorination takes place at the carbonyl carbon with substitution of the oxygen atom to give a geminal dichloride (with 2 Cl atoms on same carbon) according to the following equation: 

so we can say that acetone is converted into 2,2-dichloropropane by action of PCl₅

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Why are sodium ions good for making batteries?
Wewaii [24]

The largest advantage of sodium-ion batteries is the high natural abundance of sodium. This could make commercial production of sodium-ion batteries less expensive than lithium-ion batteries. As of 2020, sodium ion batteries have very little share of the battery market.

3 0
3 years ago
Why does the phase of the moon affect tides
Maksim231197 [3]
It's because of the gravitational forces
6 0
3 years ago
Consider the following reaction at 25 °C: 4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s) An equilibrium mixture contains 1.0 mol Fe, 1.0 × 10–3 mol
tino4ka555 [31]

Answer:

Kc = Kc = 8.0 * 10^9

Kp = 5.5 *10^5

Explanation:

Step 1: Data given

Temperature = 25.0 °C

Number of moles Fe = 1.0 moles

Number of moles O2 = 1.0 * 10^-3 moles

Number of moles Fe2O3 = 2.0 moles

Volume = 2.0 L

Step 2: The balanced equation

4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s)

Step 3: Calculate molarity

Molarity = moles / volume

[Fe] = 1.0 moles / 2.0 L

[Fe] = 0.5 M

[O2] = 0.001 moles / 2.0 L

[O2] = 0.0005 M

[Fe2O3] = 2.0 moles / 2.0 L

[Fe2O3] = 1.0 M

Step 4: Calculate Kc

Kc =1/ [O2]³

Kc = 1/0,.000000000125

Kc = 8.0 * 10^9

Step 5: Calculate Kp

Kp = Kc*(R*T)^Δn

⇒with Kc = 8.0*10^9

⇒with R = 0.08206 L*atm /mol*K

⇒with T = 298 K

⇒with Δn = -3

Kp = 8.10^9 *(0.08206 * 298)^-3

Kp = 5.5 *10^5

4 0
3 years ago
the temperature is increased, the position of equilibrium moves to the left, and the yield of ammonia decreases.
igomit [66]

Answer:i don't rlly get the question but this is what i found on the internet :/

Explanation:

When the temperature is increased, the position of equilibrium moves in the endothermic direction to reduce the temperature. ... This means that as the temperature is increased, the position of equilibrium moves to the left, and the yield of ammonia decreases.

7 0
3 years ago
In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?
dedylja [7]

<u>Answer:</u> The correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

  • <u>Reaction 1:</u>  2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 2:</u>  CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 3:</u>  HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 4:</u>  2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

  • <u>Reaction 5:</u>  SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

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3 years ago
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