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3 years ago

determine the average volume from the following volume reading: 3.00ml, 2.0ml, 2.987 x 10^-3 l, and 3.4856 ml

Chemistry

1 answer:

lesya [120]3 years ago

8 0

Answer:

The average of given values is 2.1221 ml

Explanation:

Given data:

Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml

Average value of given measurements = ?

Solution:

Formula:

Average value = sum of all measurement / total number of measurements

2.987 × 10⁻³ml = 0.002987 ml

Now we will put the values.

Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4

Average value = 8.488587 ml / 4

Average value = 2.1221 ml

The average of given values is 2.1221 ml

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3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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3 years ago

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Answer: D

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3 years ago

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Temka [501]

Answer:

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4 years ago

Sodas are not only full of sugar (124 grams/L), but they are also acidic and will dissolve your teeth, composed of hydroxyapatit

Naya [18.7K]

Answer:

The pH of the HCl solution is 3.9

Explanation:

Knowing the molar mass of HCl (36.5g/mol) it is possible to calculate the number of moles of HCl in the solution,

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