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4 years ago

6

The titration of 25.00 ml a 0.125 m hclo4 solution requires 27.07 ml of koh to reach the endpoint. what is the concentration of

the koh

1 answer:

agasfer [191]4 years ago

5 0

Answer : The concentration of the $KOH$ is, 0.115 M

Explanation :

Using dilution law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 1

$M_1$ = concentration of $HClO_4$ = 0.125 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of $HClO_4$ = 25 ml

$V_2$ = volume of NaOH = 27.07 ml

Now put all the given values in the above law, we get the concentration of the $KOH$ .

$1\times 0.125M\times 25ml=1\times M_2\times 27.07ml$

$M_2=0.115M$

Therefore, the concentration of the $KOH$ is, 0.115 M

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Answer:

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4 years ago

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3 years ago

The analysis of a hydrocarbon revealed that it was 85.7% C and 14.3% H by mass. When 1.77 g of the gas was stored in a 1.500-L f

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Answer:

The formula of hydrocarbon = $C_3H_6$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 85.7

Molar mass of C = 12.0107 g/mol

% moles of C = 85.7 / 12.0107 = 7.14

% of H = 14.3

Molar mass of H = 1.00784 g/mol

% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = $CH_2$

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (17 + 273.15) K = 290.15 K

Volume = 1.500 L

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K

⇒n = 0.0421 moles

Given that :

Amount = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0421\ moles= \frac{1.77\ g}{Molar\ mass}$

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = $C_3H_6$ </u>

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3 years ago

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PilotLPTM [1.2K]

Answer:

The answer to your question is given below.

Explanation:

Rate is simply defined as quantity per unit time. Mathematically it is represented as:

Rate = Quantity /time

Thus, we can obtain the rate as follow:

1. Quantity = 20 sandwich

Time = 10 mins

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Rate = Quantity /time

Rate = 20/10

Rate = 2 sandwich per mins

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Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 30/10

Rate = 3 sandwich per mins

3. Quantity = 40 sandwich

Time = 10 mins

Rate =?

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Rate = 40/10

Rate = 4 sandwich per mins

4. Quantity = 50 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 50/10

Rate = 5 sandwich per mins

Thus, the complete table is given as follow:

Quantity >> Unit of measure >> Rate

10 >>>>>>> 10 min >>>>>>>>>> 1

20 >>>>>>> 10 min >>>>>>>>>> 2

30 >>>>>>> 10 min >>>>>>>>>> 3

40 >>>>>>> 10 min >>>>>>>>>> 4

50 >>>>>>> 10 min >>>>>>>>>> 5

4 0

3 years ago