What is the Representative Particles in 1.00 mol of atomic nitrogen

Chemistry

1 answer:

docker41 [41]2 years ago

5 0

Answer:

a molecule.

Explanation:

You might be interested in

What is the molar concentration (in moles/L) of a solution made with 3.744 g of Mg(NO3)2 dissolved in enough water to make 50.0

Annette [7]

Answer:

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$

Explanation:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

In this case:

Mg: 24.3 g/mole
N: 14 g/mole
O: 16 g/mole

So, the molar mass of Mg(NO₃)₂ is:

Mg(NO₃)₂= 24.3 g/mole + 2*(14 g/mole + 3*16 g/mole)= 148.3 g/mole

So, if you have 3.744 g of Mg(NO₃)₂, you can apply the following rule of three: if 148.3 grams of Mg(NO₃)₂ are present in 1 mole, 3.744 grams in how many moles are present?

$moles=\frac{3.744 grams*1mole}{148.3 grams}$

moles= 0.025

Then you have:

number of moles=0.025
volume= 50 mL= 0.05 L (being 1,000 mL= 1 L)

Replacing in the definition of molarity:

$molarity=\frac{0.025 moles}{0.05 L}$

you get:

$molarity=0.5 \frac{moles}{L}$

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$

6 0

3 years ago

The change of one or more substancss into other substance

SSSSS [86.1K]

Answer:

lol what? bruh you need to explain what your talking about im in an ap class i can help

Explanation:

3 0

4 years ago

Drag the tiles to the correct boxes. Not all tiles will be used. The oxidation number of oxygen is -2, and that of chlorine is -

Evgen [1.6K]

Answer:

That oxidation number of chlorine is -1 while that of oxygen is -2

Explanation:

When it comes to manganese chloride to stabilize the chlorine Ion. Two manganese ions shall be required. Hence the formula for the compound becomes $M_nCl_2$ while the formula for manganese (iv) chloride will be $M_nCl_4$ similarly.

When oxygen and magnesium oxide is considered the Two ions of Manganese will satisfy to ions of oxygen. Which cancels and becomes $M_nO$ while for manganese (iv) oxide the formula becomes $M_nO_2$ .