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3 years ago

How to find density with radius, height, and mass.

1 answer:

4 0

The mass density ρ of an object may be found by dividing its mass M by its volume V. Sphere: V = (4/3) π R3 where R is the radius of the sphere. Cylinder: V = π R2L where R is the radius of its base and L the length of it.

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Which formula equation shows a reversible reaction?

pishuonlain [190]

Answer:2NaF is the correct one. It’s a simple combination and can be be split with relative ease

Explanation:

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3 years ago

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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

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:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

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3 years ago

What was the result of collisions between the early Earth and other, smaller protoplanets?

RoseWind [281]

The trajectory of their motion knocked the Earth into a different orbit.

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4 years ago

Are moles larger than atoms?

nataly862011 [7]

Answer:

YES BC ATMONS ARE THE SMALLEST UNIT sorry for caps

Explanation:

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3 years ago

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If the gas in the piston above has a volume of 20.0 L at a temperature of 25 C what is the volume of that gas when it is heated

dexar [7]

Answer:

15.98 L

Explanation:

First, you need to find T1, T2, V1 and V2.

T1 = 25 C = 298.15 K (25C + 273.15K)

T2 = 100 C = 373.15 K (100C + 273.15K)

V1 = 20. L

V2 = ? (we are trying to find)

Next, rearrange to fit the formula

V2 = V1 x T1 / T2

Next, fill in with our numbers

V2 = 20. L x 298.15 K / 373.15 K

Do the math and you should get...

15.98 L

- If you need more help or futher explanation please let me know. I would be glad to help!

8 0

3 years ago

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