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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 student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa
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<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

The chemical equation for the reaction of NaOH and monoprotic acid follows:

NaOH+HX\rightarrow NaX+H_2O

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = \frac{0.0225}{2}=0.01125moles

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

  • <u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

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Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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