All categories

2 years ago

How many ions are in 0.5 moles of NaCl?

Chemistry

1 answer:

noname [10]2 years ago

4 0

Explanation:

Step 1: Write out the chemical formula of the compound.

Magnesium Chloride = MgCl^2

Step 2: Convert moles into number of compounds using Avogadro's number

0.5 moles MgCl^2 ⋅(6.022⋅10^23 - 1 mole MgCl^2)= 3.011⋅ 10^23

Step 3: Determine how many Chloride ions there will be in 1 compound

There will be 2 Cl− ions in each compound (Cl^2 part)

Step 4: Multiply the number from Step 2 by the number in Step 3

3.011 ⋅ 10^23 ⋅ 2= 6.022 ⋅ 10^23

We do this because the ratio is 2 Cl− ions for every 1 compound.

In the case of finding how many ions in general (both sodium and chloride), we would multiply by 3 because there are 3 ions per 1 compound.

I hope this helped!

You might be interested in

1. A gas has a volume of 140L at 37 ºC and 620 kpa pressure. Calculate its volume at STP.

solniwko [45]

Answer:

1. Volume as STP = 755 L

2. Outside temperature = 255 K

3. Percentage yield = 70.5%

Explanation:

1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K

Using the general gas equation :

P1V1/T1 = P2V2/T2

P1 = 620 kpa

V1 = 140 L

T1 = 37°C or (273.15 + 37) K = 310.15 K

P2 = 101.3 kpa

V2 = ?

T2 = 273.15 K

V2 = P1V1T2/P2T1

V2 = 620 × 140 × 273.15 / 101.3 × 310.15

V2 = 755 L

2. Using Charles' gas law:

V1/T1 = V2/T2

V1 = 2.5 L

T1 = 290 K

V2 = 2.2 L

T2 = ?

T2 = V2T1/VI

T2 = 2.2 × 290 / 2.5

T2 = 255 K

3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu

From equation of the reaction, 2 moles of Al produces 3 moles of Cu

Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g

2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g

54 g of Al produces 190.5 g of Cu

1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu

Percentage yield = actual yield /theoretical yield × 100%

Percentage yield = 4.65/6.60 × 100%

Percentage yield = 70.5%

5 0

2 years ago

What is the purpose of the prefixes in the metric system?

Helen [10]

<span>Prefixes are used in the metric system to indicate smaller or larger measurements</span>

4 0

3 years ago

Read 2 more answers

What allowed Mendeleev to make predictions of undiscovered elements

Andrej [43]

Answer:

he predicted the properties from known elements above and belws the unknown in the same group

Explanation:

What allowed Mendeleev to make predictions of undiscovered elements

He realized that an element on this table with one known element above it and one known element below it had to have properties between the two known elements

How did Mendeleev predict gallium and germanium?

Based on gaps in the periodic table Mendeleev deduced that in these gaps belonged elements yet to be discovered. Based on other elements below and above in the same group he predicted the existence of eka-aluminum, eka-boron, and eka-silicon, later to be named gallium (Ga), scandium (Sc), and germanium (Ge).

3 0

2 years ago

Read 2 more answers

The burning of magnesium is a highly exothermic reaction. How many kilojoules of heat are released when 0. 75 mol of Mg burn in

alexandr402 [8]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

<h3>Enthalpy of a chemical reaction</h3>

The enthalpy of a chemical reaction is known as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

<h3>Heat released in this case</h3>

In this case, the balanced reaction is:

2 Mg(s) + O₂ (g) → 2 MgO(s) + 1204 kJ

This equation indicates that when 2 moles of Mg reacts with 1 mole of O₂, 1204 kJ of heat is released.

When 0.75 moles of Mg are burned, then you can apply the following rule of three: if 2 moles of Mg releases 1204 kJ of heat, 0.75 moles of Mg releases how much heat?

$heat=\frac{0.75 moles of Mgx1204 kJ}{2 moles of Mg}$