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nikitadnepr [17]
3 years ago
11

Blue litmus paper is used to determine the acidity of a compound. When litmus paper detects hydrogen ions (H+), the paper turns

red. Which solution will leave the blue litmus unchanged in color? hydrochloric acid (HCI) hydrobromic acid (HBr) B C sodium hydroxide (NaOH) sodium chloride salt (NaCl)
A. Hydrochloric acid (HCI)
B. Hydrobromic acid (HBr)
C. sodium hydroxide (NaOH)
D. sodium chloride (NaCI)​
Chemistry
1 answer:
statuscvo [17]3 years ago
8 0

Answer:

Bases Do not Change the Color of Blue Litmus.

I'd go with Option C.

Sodium Hydroxide.

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How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?
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Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

  • Volume of the solution (V): 500 mL
  • Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

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6 0
3 years ago
In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so
gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL

Putting values in above equation, we get:

1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

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

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0
3 years ago
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