Energy has both wavelength and particle-like properties, but electrons have only particle like properties, true or false

How many liters of solution would it take to make a 0.250 M solution when you have 3.52 moles of solute?

Butoxors [25]

Answer:

The volume of solution in liters required to make a 0.250 M solution from 3.52 moles of solute is 14.08 liters of solution

Explanation:

The question relates to the definition of the concentration of a solution which is the number of moles per liter (1 liter = 1 dm³) of solution

Therefore we have;

The concentration of the intended solution = 0.250 M

Therefore, the number of moles per liter of the required resolution = 0.250 moles

Therefore, the concentration of the required solution = 0.250 moles/liter

The volume in liters of the required solution that will have 3.52 moles of the solute is given as follows;

The required volume of solution = The number of moles of the solute/(The concentration of the solution)

∴ The required volume of solution = 3.52 moles/(0.250 moles/liter) = 14.08 liters

The required volume of solution to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.

Therefore the number of liters required to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.

8 0

3 years ago

Calcutale Grxn for the following equation at 25°C: <br><br> 4KClO3(s) → 3KClO4(s) KCl(s)

Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0

4 years ago

The percent yield for the reaction in which 15.6g of aluminum hydroxide is reacted in excess hydrogen chloride gas is 92.5%. Wha

Misha Larkins [42]

Answer:

24.6 g

Explanation:

4 0

3 years ago

RSB [31]

Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.

Explanation:

This is an edited book. The Harvard reference style was used in the following order:

Authors name

Year of publication

Title

Edition

Publisher

Place of publication.

Note that the title of book should be italicized with capitalization of first word.

7 0

3 years ago

If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get:

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

3 0

4 years ago