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3 years ago

When you talk together in a group during a laboratory exercise, you are engaging in

Chemistry

1 answer:

Svetlanka [38]3 years ago

7 0

Communication skills. Talking and communicating together in a group.

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Who was the first to state the concept of an atom

marin [14]

Democritus, a Greek philosopher, first developed the idea of atoms (around 460 B.C., I believe).
Hope this helps!

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4 years ago

Edith's checkerspot is a species of butterfly that lives along the west coast of North America. In recent years, scientists have

Gelneren [198K]

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3 years ago

Pls help me out!! lol

ella [17]

Answer:

i honestly dont know

Explanation:

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3 years ago

Describe how you would prepare 250 mL of a 0.707 M<br> NaNO3 solution.

timurjin [86]

Answer:

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.

Explanation:

Data given:

volume of solution = 250 mL

Convert mL to L

1 ml = 1000 L

250 ml = 250/1000 = 0.25

Molarity of solution = 0.707 M

How to prepare = ?

Solution:

Molarity:

This the term used for the concentration of the solution. It is the amount of solute in moles dissolve in 1 Liter of solution.

So we have to calculate amount of NaNO₃

So,

we have to know the number of moles of solution that will be required

Formula used for Molarity

Molarity = number of moles of solute / L of solution

Rearrange above equation

number of moles of solute = Molarity x L of solution . . . . . . . (1)

Put values in equation 1

number of moles of solute = 0.707 mol/L x 0.25 L

number of moles of solute = 0.177 mol

So we need 0.177 mol of NaNO₃ to prepare 250 mL of a 0.707 M solution.

Now convert moles to mass

Mass = no. of moles x Molar mass . . . . . . . . . . (2)

molar mass of NaNO₃ = 23 + 14 + 3(16)

molar mass of NaNO₃ = 85 g/mol

Put value in equation 2

Mass of NaNO₃= 0.177 mol x 85 g/mol

Mass of NaNO₃= 15.045 g

So now,

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.

3 0

3 years ago

Oxygen forms a number of compounds with nitrogen, many of which are quite reactive. One such compound is 36.86 % by mass N . Cal

Delicious77 [7]

Answer:

75.37 g

Explanation:

From the question given above, the following data were obtained:

Percentage of nitrogen (N) in the compound = 36.86 %

Mass of nitrogen (N) in the compound = 44 g

Mass of oxygen (O) in the compound =?

Next, we shall determine the total mass of the compound. This can be obtained as follow:

Percentage of nitrogen (N) in the compound = 36.86 %

Mass of nitrogen (N) in the compound = 44 g

Total mass of compound (T) =?

Percentage of N = mass of N/ total mass × 100

36.86 % = 44 / T

36.86 / 100 = 44 / T

Cross multiply

36.86 × T = 100 × 44

36.86 × T = 4400

Divide both side by 36.86

T = 4400 / 36.86

Total mass = 119.37 g

Next, we shall determine the percentage of oxygen in the compound. This can be obtained as follow:

Percentage of N = 36.86 %

Percentage of O =?

Percentage of O = 100 – 36.86 %

Percentage of O = 63.14 %

Finally, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Percentage of O = 63.14 %

Total mass = 119.37 g

Mass of O =.?

Percentage of O = mass of O / total mass × 100

63.14 % = Mass of O / 119.37

Cross multiply

Mass of O = 63.14 % × 119.37

Mass of O = (63.14 / 100)× 119.37

Mass of O = 75.37 g

Thus, the mass of oxygen in the compound is 75.37 g

7 0

3 years ago