What is the molarity of a solution defined as

Does sodium fluoride follow the octet rule? If not, which exception is it?

krek1111 [17]

Answer:

Sodium fluoride (NaF) does indeed follow the octet rule without any violations.

3 0

2 years ago

C3H8 + O2 → CO2 + H2O

andreyandreev [35.5K]

First, we'll balance the equation
C3H8 + 5O2 --> 3CO2 + 4H2O
Next, we know that a gas at stp occupies 22.4 liters/mole, so we can find the number of moles of oxygen gas.
15 liters O2 * (1 mole O2)/(22.4 liters O2) = 0.67 moles O2
Now we know from our balanced equation that there are 3 moles of CO2 per mole of O2, so we can find the number of moles of CO2 we produce.
0.67 moles O2 * (3 moles CO2/5 moles O2) = 0.40 moles CO2
Now, we have the number of moles of CO2, and we know that one mole occupies 22.4 liters, we can find the number of liters
.40 moles CO2 * (22.4 liters CO2)/(1 mole CO2) = 9.0 liters of CO2
The answer is D

4 0

4 years ago

If 5.85 moles of CaCO3 are used in an experiment, how many moles of carbon dioxide are created? (Be sure to include units in you

azamat

Answer:

5.85 moles of carbon dioxide are created.

Explanation:

The balanced reaction is:

CaCO₃ → CaO + CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CaCO₃: 1 mole
CaO: 1 mole
CO₂: 1 mole

Then you can apply the following rule of three: if by stoichiometry 1 mole of CaCO₃ produces 1 mole of CO₂, 5.85 moles of CaCO₃ will produce how many moles of CO₂?

$moles of CO_{2}=\frac{5.85 moles of CaCO_{3} *1mole of CO_{2} }{1 mole of CaCO_{3}}$

moles of CO₂= 5.85

5.85 moles of carbon dioxide are created.

7 0

3 years ago

A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ

navik [9.2K]

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol$

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

$0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol$

For the given chemical equation:

$2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)$

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

$0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g$

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

5 0

3 years ago

Read 2 more answers

Whats The amount of space an object takes up is known as the objects?

Ray Of Light [21]

The amount of space would be consider matter

7 0

3 years ago