What are the answers?

What is the equation for calculating gravitational potential energy on the Earth

DedPeter [7]

Gravitational potential energy = mass x g x height

(The g is measured in N/kg and is usually 9.8 or 10)

Full info in picture below:

3 0

3 years ago

.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

Strike441 [17]

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

5 0

3 years ago

Help. Help. Help. Help. Help.

Xelga [282]

Answer:

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Explanation:

7 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

4. At what temperature will 1.21 mol of oxygen (O2), exert a pressure of 1862 mmHg in a

Airida [17]

Answer:

Pressure = 1820 mmHg ... T = ( 2.39atm x 5.12L ) / ( 0.4041 moles x 0.08206atm. ... n is moles, R is the gas constant, and T is temperature in Kelvins.

Explanation:

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7 0

3 years ago