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cricket20 [7]
3 years ago
5

Bridget is in science class. Her teacher gives her two unknown substances and asks her to determine their relative pH. She place

s a piece of red litmus paper into both substances. The litmus paper turns purple when she places it into substance I. The litmus paper turns blue when she places it into substance II. Which of the following is true about the relative pH of the two substances? A. Substance I is a base and substance II is a neutral substance. B. Substance I is an acid and substance II is a base. C. Substance I is a neutral substance and substance II is a base. D. Substance I is a neutral substance and substance II is an acid.
Chemistry
2 answers:
andrezito [222]3 years ago
5 0

Answer:

c

Explanation:

the correct answer is c babes C. Substance I is a neutral substance and substance II is a base.

agasfer [191]3 years ago
4 0
Litmus paper is often used to determine the range of pH of an aqueous solution. Litmus paper specifically contains 10-15 natural dyes, in the presence of acidic solution it turns red, it turns purple when the solution is neutral and blue when dealing with basic solutions. Red litmus paper stays red for a neutral and acidic solution but changes to blue in the presence of alkaline solutions. On the other hand, blue litmus paper turns red when a solution is acidic but stays blue for neutral and alkaline solutions. Since the paper turns purple in the presence of solution 1 we know that is neutral, meanwhile, since it turns blue for the second solution we know that solution II is a base correct answer is C
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A lake has been infected by some type of new algae that is unknown. Every single day the amount of surface area that the algae t
FrozenT [24]

Answer:

It takes 86 days take to cover half of the lake

Explanation:

In the day #1, the amount of the algae is X,

In the day #2 is 2X

In the day #3 is 2*2*X = X*2²

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In the day #n the amount of the algae is X*2^(n-1)

Assuming X = 1m³. In the day 87, the area infected was:

1m³*2^(87-1)

7.74x10²⁵m³ is the total area of the lake

the half of this amount is 3.87x10²⁵m³

The time transcurred is:

3.87x10²⁵m³ = 1m³*2^(n-1)

Multiplying for 5 in each side:

ln (3.87x10²⁵) = ln (2^(n-1))

58.9175 = n-1 * 0.6931

85 = n-1

86 = n

<h3>It takes 86 days take to cover half of the lake</h3>

4 0
3 years ago
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Nana76 [90]

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6 0
3 years ago
If 17.8 grams of KOH dissolve in enough water to make a 198-gram solution, what is the concentration in percent by mass?
sleet_krkn [62]
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hope this helps!
7 0
3 years ago
Acid &amp; Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
Anuta_ua [19.1K]

<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

  • If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
  • The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

[H_3O^+]=5.012\times 10^{-5}M

The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

  • <u>For 6:</u>

We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0
3 years ago
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