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Roman55 [17]
3 years ago
14

Please help with this lab !

Chemistry
1 answer:
vekshin13 years ago
4 0
To get out the iron filings put everything in to a container and then put the magnet to the side to attract them. Pull them out of the container. Then, I don’t know how to separate the sand and salt. Maybe evaporate the salt? I joke you can figure it out. Best of luck! :)
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Each element is____than____element.
Ahat [919]
Each element is diffrent than another element 
6 0
2 years ago
How many molecules are in 18moles of CH.​
-Dominant- [34]

Answer:

1.08 x 10²⁵molecules

Explanation:

From the mole concept we know that ;

      1 mole of any substance contains 6.02 x 10²³ molecules

This number is the Avogadro's number.

 So;

  18 mole of CH will contain:

       Number of molecules of CH  = 18 x 6.02 x 10²³ = 1.08 x 10²⁵molecules

The number of molecules is therefore 1.08 x 10²⁵molecules

8 0
2 years ago
Bond order of n2<br><br>bond order of N2 ​
snow_lady [41]

Answer:

The MO method for N2+ gives the bond order equal to 2.5. But first, we look at the diagram of molecular orbitals for N2 (the bond order for the nitrogen molecule is 3). the N2+ molecule). That is, the bond order for N2+ is 2.5.

4 0
2 years ago
Does anyone know how to do this ??
lord [1]

Answer:

Look at your periodic table

Explanation:

It tells you everything like Potasium 1 and 1 ox and 1 hydrogen

7 0
3 years ago
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0
2 years ago
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