Presence of unlimited number of elements in the periodic table The inability of atoms to align perfectly with other atoms The ability of all elements to react with every other element All elements being equally reactive.

6 0

3 years ago

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ladessa [460]

Answer:

The molarity of the solution is 1.1 $\frac{moles}{liter}$

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

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

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

Molarity is expressed in units $\frac{moles}{liter}$

In this case

number of moles of solute= 0.564 moles
volume= 0.510 L

Replacing:

$molarity=\frac{0.564 moles}{0.510 L}$

Solving:

molarity= 1.1 $\frac{moles}{liter}$

The molarity of the solution is 1.1 $\frac{moles}{liter}$

4 0

3 years ago

Se combustionan 2 g de magnesio en Aire, obteniéndose 3,28 g de óxido de magnesio (MgO)¿Qué masa de oxígeno se consumen en la re

konstantin123 [22]

Answer:

1.30 g

Explanation:

La reacción que toma lugar es:

2Mg + O₂ → 2MgO

Primero convertimos 3.28 g de MgO en moles, usando su masa molar:

3.28 g ÷ 40.3 g/mol = 0.081 mol MgO

Después convertimos 0.081 moles de MgO en moles de O₂, usando los coeficientes estequiométricos:

0.081 mol MgO * $\frac{1molO_2}{2molMgO}$ = 0.0407 mol O₂

Finalmente convertimos 0.0407 moles de O₂ en gramos:

0.0407 mol O₂ * 32 g/mol = 1.30 g

5 0

3 years ago

How many moles of AgCl will be produced from 60.0 g of AgNO3, assuming NaCl is available in excess?

Ivenika [448]

Answer: For this homework question, you’ll need to look up the molar mass of AgNO₃. Use this to calculate the number of moles in the sample.

Then note that silver is (quite usually) univalent, so there will be as many moles of AgCl as there were of AgNO₃.

Ask your instructor or tutor for more in-depth training, if need be. Good luck!

Explanation:

4 0

3 years ago