The quantity (current × voltage) is:<br><br><br><br> please hurry guys, I need helps

I NEED HELP NOW!!!!!!!!!!!

Rashid [163]

A: the ball in frame A had the highest velocity, and the ball in frame B has the highest kinetic energy

3 0

3 years ago

If the weight of displaced liquid and the weight of an object are the same, the object will:

s344n2d4d5 [400]

Answer:

Float

Explanation:

If the weight of liquid displaced and the weight of an object are the same, the object will float in the liquid.

From Archimedes principle, when an object is immersed in fluid, a force called upthrust supports it and it equal to the weight of the liquid displaced.

When in a liquid, the weight of the liquid displaced is the same as that of the of the object, it will float and not sink.

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

Which of the following describes a nonmetal

anastassius [24]

Answer:

b is the answer

Explanation:

non metals are not shiny, brittle, unmalleable, and are poor conductors of thermal energy and electrical current.

4 0

3 years ago

I need help on this question on standard enthalpy of formation

forsale [732]

Answer:

B: Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)

Explanation:

We are looking for enthalpy of formation, so we want to see reactance in their natural standard form.

Thus, we want to see the reactance of Na, Cl2 and O2.

The only option that has the correct form of Na, Cl2 and O2 is B.

Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)

7 0

2 years ago

The quantity (current × voltage) is: please hurry guys, I need helps