Methane is a gas that easily catches fire. This statement is describing a ____

Calculate how many grams would be required to prepare 600. 0 ml of 0. 150 m of naf? molar mass of naf is 41. 9 g/mo

marta [7]

Considering the definition of molarity and molar mass, the mass of NaF required is 3.771 grams.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

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

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

Molarity is expressed in units $\frac{moles}{liters}$ .

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3 /><h3>Mass of NaF required</h3>

In this case, you know:

molarity= 0.15 M= 0.15 $\frac{moles}{L}$
number of moles= ?
volume= 600 mL= 0.6 L

Replacing in the definition of molarity:

$0.15\frac{moles}{L} =\frac{number of moles}{0.6 L}$

Solving:

0.15 $\frac{moles}{L}$ × 0.6 L= number of moles

<u><em>0.09 moles= number of moles</em></u>

The molar mass of NaF is 41. 9 g/mol. So, you can apply the following rule of three: If by definition of molar mass 1 mole of the compound contains 41.9 grams, 0.09 moles of the compound contains how much mass?

$mass=\frac{0.09 molesx41.9 grams}{1 mole}$

<u><em>mass= 3.771 grams</em></u>

Finally, the mass of NaF required is 3.771 grams.

Learn more about

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4 0

2 years ago

Compound A has a partition coefficient (K) of 7 when comparing its solubility in CH2Cl2 to water ( K=11, [solubility of A in g/m

Hitman42 [59]

Answer:

The equation which will tell us how much of A that is mis inmthe water layer after partitioning is: 7 = (17 - x) g / 150 mL ÷ x g /100 mL

Explanation:

A partition coefficient is the ratio of the concentration of a substance in one solvent phase to the concentration in a second solvent phase when the two concentrations are at equilibrium. Usually the two phases are an organic phase and an aqueous phase. Thus, the partition coefficient K, of a compound is the ratio of the compound's concentration in the organic layer compared to the aqueous layer.

K = C₁/C₂ at equilibrium

In the compound A given, CH₂Cl₂ is the organic phase while water is the aqueous phase

Amount of A that is partitioned in between dichloromethane, CH₂Cl₂ and water, H₂O is 17.0 g

Let the amount of A that is dissolved in water be x g

Solubility of A in water given in g/mL = (x / 100) g/ml

Amount of A dissolved in dichloromethane, CH₂Cl₂ = (17 - x) g

Solubility of A in dichloromethane, CH₂Cl₂ given in g/mL = (17 - x/150) g/mL

Since the partition coefficient, K of compound A when comparing its solubility in CH₂Cl₂ to water is 7, that is;

K = [solubility of A in g/ml in CH₂Cl₂] / [solubility of A in g/ml in H2O] = 7

The equation for the amount of A in the water layer is given as follows:

7 = (17 - x) g / 150 mL ÷ x g /100 mL

Solving for x

7 = (17 -x) × 100 / 150x

7 × 150x = (17 - x) × 100

1050x = 1700 - 100x

1150x = 1700

x = 1700/1150

x = 1.48 g

5 0

3 years ago

What is equal to the number of particles in 36 grams of carbon-12?

HACTEHA [7]

Answer:

3 moles

Explanation:

To solve this problem we will use the Avogadro numbers.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms, ions or molecules in one mole of substance. According to this,

1.008 g of hydrogen = 1 mole = 6.022×10²³ atoms.

18 g water = 1 mole = 6.022×10²³ molecules

we are given 36 g of C-12. So,

12 g of C-12 = 1 mole

24 g of C-12 = 2 mole

36 g of C-12 = 3 mole

So 3 moles of C-12 equals to the number of particles in 36 g of C-12.

4 0

3 years ago

UNIT TEST

Aleksandr-060686 [28]

Answer:

Six

Explanation:

I just took it and that is the correct answer.

8 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

Methane is a gas that easily catches fire. This statement is describing a _____.