All categories

3 years ago

Calculate the molarity of the two solutions.

Chemistry

1 answer:

beks73 [17]3 years ago

8 0

Answer:

A = Molarity = 0.22 M

B = Molarity = 0.36 M

Explanation:

Given data:

For first solution:

number of moles = 0.550 mol

Volume of solution = 2.50 L

Molarity = ?

Molarity:

Formula:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.550 mol / 2.50 L

Molarity = 0.22 M

For second solution:

Mass of NaCl = 15.7 g

Volume of solution = 709 mL or 709/1000 = 0.709 L

Molarity = ?

Solution:

Number of moles = mass / molar mass

Number of moles = 14.7 g/ 58.44 g/mol

Number of moles = 0.252 mol

Molarity:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.252 mol / 0.709 L

Molarity = 0.36 M

You might be interested in

An acetylene tank has a volume of 390.0 L. It is stored at a temperature of 23.5 °C and has a

xeze [42]

Considering the ideal gas law, there are 279.42 moles of acetylene in the tank.

<h3>Definition of ideal gas</h3>

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

<h3>Ideal gas law</h3>

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Moles of acetylene</h3>

In this case, you know:

P= 1765 kPa= 17.4192 atm (being 101.325 kPa= 1 atm)
V= 390 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 23.5 °C= 296.5 K (being 0 °C= 273 K)

Replacing in the ideal gas law:

17.4192 atm× 390 L = n×0.082 $\frac{atmL}{molK}$ × 296.5 K

Solving:

$n=\frac{17.4192 atmx 390 L}{0.082 \frac{atmL}{molK}x296.5 L}$

<u><em>n= 279.42 moles</em></u>

Finally, there are 279.42 moles of acetylene in the tank.

Learn more about ideal gas law:

brainly.com/question/4147359

4 0

2 years ago

Describe why the overgrowth of algae in eutrophication is a bad thing.

Tanya [424]

Answer:

The overgrowth of algae consumes oxygen and blocks sunlight from underwater plants. When the algae eventually dies, the oxygen in the water is consumed. The lack of oxygen makes it impossible for aquatic life to survive

Explanation:

hope this helps have a good rest of your afternoon :) ❤

7 0

3 years ago

Read 2 more answers

Consider the following reaction: 2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s)

Mandarinka [93]

Taking into account the reaction stoichiometry, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Bi + 3 Cl₂ → 2 BiCl₃

By reaction stoichiometry (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:

Bi: 2 moles
Cl₂: 3 moles
BiCl₃: 2 moles

The molar mass of the compounds is:

Bi: 209 g/mole
Cl₂: 70.90 g/mole
BiCl₃: 315.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Bi: 2 moles ×209 g/mole=418 g

Cl₂: 3 moles ×70.90 g/mole= 212.7 g

BiCl₃: 2 moles ×315.45 g/mole= 630.9 g

<h3>Mass of Cl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 630.9 grams of BiCl₃ are formed by 212.7 grams of Cl₂, 3.32 grams of BiCl₃ are formed by how much mass of Cl₂?

$mass of Cl_{2} =\frac{3.32 grams of BiCl_{3}x212.7 grams of Cl_{2} }{630.9 grams of BiCl_{3}}$

<u><em>mass of Cl₂= 1.119 grams</em></u>

Finally, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

5 0

2 years ago

Air pressures for thunderstorms

aliina [53]

Answer:

Explanation:

The typical baseline reading is done at sea level, where 14.7 pounds of pressure per square inch of air will raise the mercury level to 29.53 inches. Air pressure falls as the elevation rises

please give brainlest tryin to grind

7 0

3 years ago

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an

mafiozo [28]

Answer:

$\rm K$ : approximately $22.0\%$ .

$\rm Cl$ : approximately $19.9\%$ .

$\rm O$ : approximately $27.0\%$ .

$\rm Al$ : $31.0\%$ .

Explanation:

Consider a $100\; \rm g$ sample. There would be $69.0\% \times 100\; \rm g = 69.0\; \rm g$ of $\rm KClO_{3}$ and $100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g$ of $\rm Al$ .

Out of that $69.0\; \rm g$ of $\rm KClO_{3}$ :

$31.9\%$ (by mass) would be $\rm K$ : $31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g$ .

$28.9\%$ (by mass) would be $\rm Cl$ : $28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g$ .

$39.2\%$ (by mass) would be $\rm O$ : $39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g$ .

Overall, the composition (by mass) of each element in this mixture would be:

$\rm K$ :

$\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}$ .

$\rm Cl$ :

$\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}$ .

$\rm O$ :

$\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}$ .

$\rm Al$ :

$\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}$ .

7 0

3 years ago