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Vitek1552 [10]
3 years ago
14

what type of wave uses thioglycolic acid or its derivatives with ammonia and procceses the hair without heat?

Chemistry
1 answer:
GaryK [48]3 years ago
6 0

Answer:

Cold Waves

Explanation:

Cold Waves -

This type of wave are used to curl the hairs without the application of any type of heat and along with it , thioglycolic acid or any of its derivatives are used along with ammonia , during the process of curling .

The cold wave use some small or medium size plastic curlers and obtaining a result of medium curls which are more prominent when they are wet .

For better results hair products like mousse , setting lotions are applied .

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Net ionic of ammonium sulfide added to iron (ll) chloride
meriva

Answer:

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)  

Step-by-step explanation:

Molecular Equation:

(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)

Ionic equation :

2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)

Net ionic equation :

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)

4 0
3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

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Answer: the answer is white dwarf

Explanation:

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4 moles of hydrogen chloride (note that it is in the gaseous phase, otherwise it would be hydrochloric acid) react with 1 mole of oxygen gas to form 2 moles of chlorine gas and 2 moles of liquid water.

To conform with  the law of conservation of mass, the equation must be balanced, this means that there must be the same number of each type of atom  on both sides of the arrow.

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Answer:

no im made outta atoms and stuff like that

Explanation:

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