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3 years ago

What is the total volume of solution that was dispensed from this burette? before after

Chemistry

2 answers:

Lemur [1.5K]3 years ago

8 0

Answer : The correct option is, (C) 0.6 ml

Explanation : Given,

Initial volume of solution in the burette = 32.3 ml

Final volume of solution in the burette = 32.9 ml

As the liquid is colorless. So, we are taking the lower meniscus for the correct reading of the solution.

Now we have to calculate the total volume of solution that was dispensed from this burette.

$\text{Total volume of solution dispensed}=\text{Final volume of solution in the burette}-\text{Initial volume of solution in the burette}$

Now put all the values in this formula, we get the total volume of solution that was dispensed from this burette.

$\text{Total volume of solution dispensed}=32.9ml-32.3ml=0.6ml$

Therefore, the the total volume of solution that dispensed from this burette was, 0.6 ml

monitta3 years ago

6 0

Yeah answer should be .6 ml

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A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

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Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

<u>Step by Step by explanation:</u>

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

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Result:

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3 years ago

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)

lesya692 [45]

Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.

Explanation:

Given: Current = 62.0 A

Time = 23.0 sec

Formula used to calculate charge is as follows.

$Q = I \times t$

where,

Q = charge

I = current

t = time

Substitute the values into above formula as follows.

$Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C$

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

$Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol$

The oxidation state of Pb in $PbSO_{4}$ is 2. So, moles deposited by Pb is as follows.

$Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol$

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g$

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.

3 0

3 years ago