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3 years ago

How many liters of a 0.5M sodium hydroxide solution would contain 2 mols of solute

Chemistry

2 answers:

Airida [17]3 years ago

4 0

Molarity = mol/L
(0.5M) = (2mol)/L
(2mol)/(0.5M) = L
4 liters

icang [17]3 years ago

3 0

Answer: The volume of sodium hydroxide solution is 4L.

Explanation:

Molarity is defined as the number of moles present in one liter of solution. Equation used to calculate molarity of the solution is:

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

We are given:

Moles of sodium hydroxide = 2 moles

Molarity of the solution = 0.5 moles/ L

Putting values in above equation, we get:

$0.5mol/L=\frac{2mol}{\text{Volume of solution}}\\\\\text{Volume of the solution}=4L$

Hence, the volume of the sodium hydroxide solution is 4 L.

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In order to balance an equation, we apply the principle of conservation of mass, which states that mass can neither be created nor destroyed. Therefore, the mass of an element before and after a reaction remains constant. Here, the balanced equation becomes:
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3 years ago

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What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso

Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

$pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247$

The pH of the buffer can be known as

$pH = pK_{a} + log[\frac{[A-]}{[HA]}}]$

The concentration of $[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\$

Similarly, the concentration of [HA] = $\frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404$

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

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4 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago

When 10.0 grams of ch4 reacts completely with 40.0 grams of o2 such that there are no reactants left over, 27.5 grams of carbon

Zina [86]

The balanced equation for the reaction is :-

CH₄(g) + 2O₂(g) ---------> CO₂(g) + 2H₂O(l)

Molar mass of CH₄ = 16 g/mole

Molar mass of O₂ = 32 g/mole

Molar mass of H₂O = 18 g/mole

Molar mass of CO₂ = 44 g/mole

Now we calculate the number of moles of reactants,

moles of CH₄ = mass of CH₄/molar mass of CH₄ = 10/16 = 0.625

Moles of O₂ = mass of O₂/molar mass of O₂ = 40/32 = 1.25

Now, as per the balanced reaction, for complete reaction to occur, one mole of CH₄ require 2 moles of O₂

Thus, 0.625 moles of CH₄ requires 1.25 moles of O₂

both CH₄ and O₂ are present in the exact required quantity.

Hence, moles of CO₂ formed = moles of CH₄ reacted = 0.625

Mass of CO₂ formed = moles of CO₂ x Molar mass of CO₂ = 0.625 x 44 = 27.5g

Moles of H₂O formed = moles of O₂ reacted = 1.25g

Thus, mass of H₂O formed = moles of H₂O x molar mass of H₂O = 1.25 x 18 = 22.5g

8 0

3 years ago