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2 years ago

A lead cylinder has a mass of 540 grams and a density of 2.70 g/ml. What is its volume

Chemistry

2 answers:

ch4aika [34]2 years ago

3 0

Volume = Mass / Density

Volume = 540g / 2.70 g/ml

Volume = 200 ml

Solnce55 [7]2 years ago

3 0

Answer:

Volume of the cylinder is, $V=200\ ml$

Explanation:

It is given that,

Mass of the cylinder, m = 540 grams

Density of the cylinder, $d=2.7\ g/ml$

Let V is the volume of the material. We know that the mass per unit volume of any material is called its density. The mathematical relation is given by :

$d=\dfrac{m}{V}$

$V=\dfrac{m}{d}$

$V=\dfrac{540\ g}{2.7\ g/ml}$

$V=200\ ml$

So, the volume of the cylinder is $200\ ml$ . Hence, this is the required solution.

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Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M

Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

$Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V$

Reduction half reaction: $Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V$

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: $2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.45-(-1.66)=1.21V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

$E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V$

5 0

2 years ago

If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?

kap26 [50]

Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!

6 0

2 years ago

Read 2 more answers

A poisoned pill contains 0.00048 moles of KCN. How many molecules are in this sample?

mario62 [17]

Answer:

$2.89 \times {10}^{20} \: \: molecules$