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irga5000 [103]
3 years ago
6

Balance the following reactions: 1. PbO2 → PbO + O2

Chemistry
1 answer:
Andre45 [30]3 years ago
3 0

Explanation:

2 moles of lead oxide ->

2PbO2 -> 2PbO + O2

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Balance the equation<br> N2O5 + H2 -&gt; NH3 + H2O
Lubov Fominskaja [6]

Answer:

I hope this is it. I'm not really sure.

4 0
3 years ago
Balance the following reaction: AL2O3 + HCI ⇒ ALCI3 + H2O
Anastaziya [24]
To balance the following chemical equation, make a tally or a count of each of the atoms on both sides of the reaction, and make sure that those atoms are equal on both the reactant and product side.

AL2O3 + HCl => ALCl3 + H2O

Left side. Right side
AL = 2. AL = 1
O = 3 Cl = 3
H = 1. H = 2
Cl = 1. O = 1

First balance the metal atoms, aluminum, then hydrogen and then oxygen.

Balanced equation :
AL2O3 + 6HCl => 2ALCl3 + 3H2O.

Left side. Right side
AL = 2 AL = 2
O = 3 Cl = 6
H = 6 H = 6
Cl = 6 O = 3.

4 0
3 years ago
Benzene can be nitrated with a mixture of nitric and sulfuric acids. How do we do that?
Montano1993 [528]

Answer:

We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C

Explanation:

Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C

The nitration of benzene takes place through electrophilic substitution reaction

In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene

These electrophile is generated when nitric acid is treated with sulphuric acid

As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3

The reactions regarding the nitration of benzene is present in the file attached

5 0
3 years ago
How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat
stira [4]

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

8 0
3 years ago
Someone please help me
Vladimir79 [104]

Answer:reflection

Explanation:

6 0
3 years ago
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