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2 years ago

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50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s

olution was 20g/dm3. Work out the concentration of the acid in grams per litre if it took 25cm3 of acid to completely neutralise the alkali. The relative molecular mass of sulfuric acid is 98.

1 answer:

miskamm [114]2 years ago

4 0

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

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A branched alkane has HIGHER boiling point relative to the isomeric linear alkane. There are STRONGER london force interactions in the branched alkane.

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2 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

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Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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2 years ago

William adds two values, following the rules for using significant figures in computations. He should write the sum of these two

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<span>when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.
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2 years ago

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Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(c) $Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

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2 years ago

ANSWER FAST PLEASE!!

Svetach [21]

Answer:D

Explanation:

Hope I helped

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2 years ago