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3 years ago

In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution

Chemistry

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

The molarity is "0.050 M".

Explanation:

The given values are:

M1 = 0.100 M

M2 = ?

V1 = 25.0 mL

V2 = 50.0 mL

As we know,

⇒ $M1\times V1=M2\times V2$

Or,

⇒ $M2=\frac{M1\times V1}{V2}$

By putting the values, we get

$=\frac{0.100\times 25}{50}$

$=\frac{2.5}{50}$

$=0.05 \ M$

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Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to o

Viefleur [7K]

Answer:

$P_2=404 kPa$

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

$P_1V_1=P_2V_2$

Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:

$V_2=\frac{1}{4} V_1$

We can compute the new pressure:

$P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa$

Which means the pressure is increased by a factor of four.

Regards.

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3 years ago

For the reaction below , label each reactant as an electron pair acceptor or electron pair donor and as a Lewis acid or a Lewis

JulsSmile [24]

Since the reaction shown in the question is an acid - base reaction in the Lewis sense; the Lewis acid here is AlCl3 while the Lewis base here is Cl^- .

<h3>What is a Lewis acid?</h3>

A Lewis acid is a substance that accepts electron pair while a Lewis base donates an electron pair.

Now consider the given reaction; AlCl3 +Cl^- ------> AlCl 4 ^-. The Lewis acid here is AlCl3 while the Lewis base here is Cl^- .

Learn more about acid - base reaction: brainly.com/question/14356798

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2 years ago

Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

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4 years ago

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3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago