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3 years ago

10

A high jumper jumps 2.09m. If the jumper has a mass of 72kg, what is his gravitational potential energy at the highest point of

the jump?

1 answer:

Furkat [3]3 years ago

4 0

Si Ko alam hehehehe sorrry men

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What is the oxidation half-reaction for 2Mg + O2 + 2Mgo?

miv72 [106K]

Answer:

B => Mg° => Mg⁺² + 2e⁻

Explanation:

2Mg° + O₂ => 2MgO rewrite => Mg° + 1/2O₂ => 2MgO

Oxidation Half Rxn => Mg° => Mg⁺² + 2e⁻

Reduction Half Rxn => 1/2O₂ + 2e⁻ => 2O⁻

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FYI Note => For half-reactions one always finds the electrons listed on the product side of the reaction for oxidation reactions and for reduction reactions on the reactant side.

5 0

3 years ago

What is the calculated value of the cell potential at 298K for an

Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

<u>Answer:</u> The cell potential of the given electrochemical cell is 0.273 V

<u>Explanation:</u>

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

<u>Oxidation half reaction:</u> $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

<u>Reduction half reaction:</u> $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

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:

$E^o_{cell}=0.0-(-0.14)=0.14V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V

4 0

3 years ago

What color is the 2,6-dichloroindophenol solution in acidic solutions?

ICE Princess25 [194]

DCPIP is a redox dye that starts off blue but will turn reddish-pink in acidic solution.

This chemical is often used in general chemistry experiments with vitamin C (ascorbic acid), which is a good reducing agent. The DCPIP become colorless when it is reduced (remember, it's blue when oxidized).

7 0

3 years ago

A sample of a substance was determined and contained 1.0g of mg and 1.75g of o. what is the empirical formula? (10 pts)

Juliette [100K]

<span>Total mass = 2.75g Mass % of Mg = 1.0g x 100/2.75g = 36.36 % Mass % of O = 1.75g x 100/2.75g = 63.64 % Mol of Mg = 36.36/24 = 1.515 Mol of O = 63.64/16 = 3.977 Ratio of Mol of Mg and O in the substance = (1.515 : 3.977) x 2 = 3 : 8 The empirical formula of substance is Mg3O8</span>

7 0

3 years ago

Chase had determined the boiling point of an unknown liquid to be 45.2 ⁰C when the correct boiling point is 44.32 ⁰C. What is th

Nataly_w [17]

Answer:

The boiling point of the liquid is 47.368°C

Explanation:

4 0

2 years ago