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Ostrovityanka [42]
3 years ago
8

Can someone please help as soon as possible

Chemistry
1 answer:
Kobotan [32]3 years ago
4 0

Answer:

Explanation:

#1 2HCL and 2H20

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What two things combine with oxygen to form ground-level ozone?
STALIN [3.7K]

Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

-

4 0
3 years ago
Arrange the compounds by their reactivity toward electrophilic aromatic substitution.
swat32

Answer:

The order of reactivity towards electrophilic susbtitution is shown below:

a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene

b. p-cresol>p-xylene>toluene>benzene

c.Phenol>propylbenzene>benzene>benzoic acid

d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene

Explanation:

Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.

For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.

The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.

Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.

4 0
3 years ago
The first strings of holiday lights were manufactured as series circuits. Which explains why this method is no longer used? A. I
user100 [1]
B because once the circuit is burned the lights will all go of
3 0
3 years ago
How much heat is required to vaporize 25g of water at 100*c
ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

           L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

4 0
3 years ago
Boron has two isotopes, Boron 10 which has a mass of 10.0129 amu and Boron-11 with a mass of 11.0093 amu. B-10 occurs 84.74% of
podryga [215]

The average atomic mass of Boron: 10.431 amu

<h3>Further explanation   </h3>

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.  

So Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu  

Mass atom X = mass isotope 1 . % + mass isotope 2.%  + ...

The average atomic mass of boron :

\tt avg~mass=0.8474\times 10.0129+0.1526\times 11.0093\\\\avg~mass=10.431~amu

3 0
3 years ago
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