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Anastasy [175]
3 years ago
10

A student performs a titration procedure by adding NaOH to a solution that contains H+ ions. She added the indicator, phenolphth

alein, to the flask containing the H+ ions. What was the purpose of adding the indicator to the flask?
A. The indicator adds more H+ ions to the solution which allows the titration procedure to proceed faster.


B. The indicator is expected to turn pink when the end point of the neutralization reaction occurs.


C. The indicator was used to determine when the amount of H+ ions was greater than the concentration of OH- ions.


D. Because the indicator has the same pH as the NaOH, it is used to determine the pH of the titrant.
Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0
Indicators are substances that produce colour changes to indicate the presence or absence of a threshold concentration of a particular chemical specie. In neutralization reaction, indicators produce colour changes on the basis of the pH of the solution. In acid base titration, indicators are used to indicate the end point of the neutralization reaction. The end point of acid base titration is the equivalence point when the moles of the of the standard solution is equal to the moles of the unknown solution. Thus, the correct option is B. Note that the solution that contain the H+ ions is the acid.
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Answer:

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Explanation:

4 0
4 years ago
Is pOH=4.3 an acid or a base
AfilCa [17]

Answer:

a base

Explanation:

first convert the pOH to pH, that way it will

be easy to decide.

formula:pOH+pH=14

pH=14-4.3=9.7

high pH tells us it's a base

6 0
3 years ago
Check all that apply.
White raven [17]

Answer:

4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

option 1,2,3 and 4 are correct. Option 5 is not correct

Explanation:

Step 1: Data given

Oxidation number of H = +1

Oxidation number of O = -2

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: The oxidation numbers

-4 for C in CH4

 ⇒ Oxidation number of H = +1

    ⇒ 4x +1 = +4

C has an oxidation number of -4

This is correct.

+4 for C in CO2

 ⇒ Oxidation number of O = -2

    ⇒ 2x -2 = -4

C has an oxidation number of +4

This is correct.

-2 for O in all substances

⇒ this is correct, the oxidation number of O is always -2 (except in H2O2 and Na2O2)

This is correct.

+1 for H in both CH4 and H2O

⇒ this is correct, the oxidation number of H is always +1 (except in metal hydrides).

This is correct.

+4 for O in H2O

 ⇒ Oxidation number of H = +1

    ⇒ 2x +1 = +2

The oxidation number of O is -2

This is not correct

5 0
3 years ago
Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
3 years ago
Anyone know how to do this???
IgorC [24]

Answer:

nope

Explanation:

cuz i cant do math

8 0
3 years ago
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