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Ostrovityanka [42]
3 years ago
6

Which of the following is not an indication of a chemical change?

Chemistry
1 answer:
Sauron [17]3 years ago
5 0

Answer:

C. precipitate formation.

Explanation:

options C is correct answer.

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B. NaOH

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Which two types of food are homogeneous mixtures
Anni [7]

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6 0
2 years ago
How many moles of potassium nitrate are required to make 550 mL of a 2.1M solution?
stira [4]

Answer:

1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes.

Molarity is a unit of concentration of a solution and indicates the amount of moles of solute that appear dissolved in each liter of the mixture. In other words, the Molarity (M) or Molar Concentration is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

Molatity( M)=\frac{number of moles of solute}{Volume}

Molarity is expressed in units (\frac{moles}{liter}).

In this case:

  • Molarity= 2.1 M
  • number of moles of solute= ?
  • Volume= 550 mL= 0.550 L (being 1L=1000 mL)

Replacing:

2.1 M= 2.1 \frac{moles}{liter} =\frac{number of moles of solute}{0.550 liters}

Solving:

number of moles of solute= 2.1 M* 0.550 L

number of moles of solute= 1.155 moles

1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.

4 0
3 years ago
What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?
Andreyy89

Answer: The volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

Explanation:

To calculate the volume 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 H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Na_2CO_3

We are given:

n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL

Thus volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

5 0
3 years ago
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