Separating mixtures: filtration

Distinguish and give example for homogeneous and heterogeneous mixture

Nataliya [291]

Answer:

Through combining two or more substances, a mixture is produced. A homogeneous solution tends to be identical, no matter how you sample it. Homogeneous mixtures are sources of water, saline solution, some alloys, and bitumen. Sand, oil and water, and chicken noodle soup are examples of heterogeneous mixtures.

Explanation:

6 0

3 years ago

What is the land like where mesosaurus fossils are found?

Veronika [31]

Answer:

They are found in South America and African Plates. Earths outer layer is made out of solid rock. These fossils are sometimes discovered closer to the mantle; the mantle is between the crust and the earths super-heated core!

Explanation:

Copy the answer and im sure you'll get it right!

Have a wonderful day/night and believe in yourself!

3 0

1 year ago

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Fill in the blanks....

sveta [45]

Iron 1 chlorine 3 for number what I suggest you do is look at the periodic table online and then look at the numbers and see how many are there

3 0

2 years ago

What is the coefficient for oxygen in the balanced equation? C 5H 12 + ? O2 → ? CO2 + ? H2O. 2 4 5 6 8

Temka [501]

Answer:

8

Explanation:

When you balance the entire equation, you should get:

C5H12 + 8O2 ---> 5CO2 + 6H2O

6 0

3 years ago

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In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago

Separating mixtures: filtration​

Separating mixtures: filtration