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N76 [4]
3 years ago
15

Changing subscripts is one of the steps of balancing a chemical equation. True or False

Chemistry
2 answers:
VladimirAG [237]3 years ago
8 0
The answer is false
have a good night !!!
Lostsunrise [7]3 years ago
3 0

Answer:

False

Explanation:

Changing the coefficients is one of the steps of balancing a chemical equation. Changing the subscript changes the compounds being used, while changing the coefficient changes the amount of each compound being used.

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Formula for all possible isomers of hexane​
Darya [45]

Answer:

1) n-Hexane. 2) 2-Methyl pentane(IUPAC name) or Isohexane(common name). 3) 2,2-Dimethylbutane(IUPAC name) or Neohexane(common name). 4) 3-Methylpentane

Explanation:

8 0
3 years ago
The mass of any amount of a solid, liquid, or gas relative to its volume is called its ____________.
zheka24 [161]
Eazy, it’s called density

I’m only in 8th grade science you know
8 0
3 years ago
Watch the animation and observe the titration process of a standard 0.100 M sodium hydroxide solution with 50.0 \rm ml of a 0.10
yuradex [85]

Answer:

Answers are in the explanation

Explanation:

a. The statments are:

The titration process is based on a chemical reaction.  <em>TRUE. </em>A titration is the chemical reaction between an acid and a base.

The pH of the solution at the equivalence point is 7.  <em>TRUE. </em>In a titration of strong acid with strong base the equivalence point is 7

At the endpoint, the pH of solution is 10 and the color of solution is pink.  <em>FALSE</em>. The endpoint in a strong acid - strong base titrationmust be near to equivalence point at pH = 7.

At the beginning of the titration process, the pH of the solution increases rapidly.  <em>FALSE. </em>At beginning of a titration, the pH of the solution increases slowly.

The chemical reaction involved in an acid-base titration is a neutralization reaction.  <em>TRUE. </em>In the reaction, you are neutralizing an acid (HCl) with a base (NaOH)

Before any base is added to the solution, the pH of the solution is high. <em>FALSE. </em>The addition of a base increases pH that is, in the beginning, low.

In a titration process, the endpoint is reached before the equivalence point.  <em>FALSE. </em>In a titration process, the endpoint is reached in the equivalence point

The pH of the solution changes very slowly at the equivalence point. <em>FALSE. </em>At the equivalence point, the pH changes rapidly.

b. For the reaction

NaOH + HCl → H₂O + NaCl

As the reaction is 1:1 and molarity of both solutions are the same, additions < 100mL of NaOH will stay before the equivalence point, equivalence point will be in 100mL and additions > 100mL of NaOH will stay after the equivalence point.

The conditions are:

10.0mL of 1.00 M NaOH before the equivalence point

150 mL of 1.00 M NaOH after the equivalence point

5.00 mL of 1.00 M NaOH before the equivalence point

50.0 mL of 1.00 M NaOH  before the equivalence point

200 mL of 1.00 M NaOH after the equivalence point

I hope it helps!

7 0
3 years ago
Read 2 more answers
Which of the following objects would require the use of a large-scale model A. An atom B. A car engine C. A frog's heart D. The
levacccp [35]

Answer:

A. An atom

Explanation:

A large scale model is one that is bigger than the object itself. These are needed to visualise structures and behaviours that we cannot normally see. A large scale model is needed for small objects, such as an atom. Larger objects do not need<em> even bigger</em> models

4 0
3 years ago
Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu
MAVERICK [17]

Answer : The initial temperature of system 2 is, 19.415^oC

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

-q_1=q_2

m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)

The mass remains same.

where,

C_1 = heat capacity of system 1 = 19.9 J/mole.K

C_2 = heat capacity of system 2 = 28.2 J/mole.K

T_f = final temperature of system = 30^oC=273+30=303K

T_1 = initial temperature of system 1 = 45^oC=273+45=318K

T_2 = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K

T_2=292.415K

T_2=292.415-273=19.415^oC

Therefore, the initial temperature of system 2 is, 19.415^oC

8 0
4 years ago
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