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2 years ago

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You apply 100.0 of force to a machine. The machine applies 20.0 N of force to the load. What is the mechanical advantage of the

machine.

1 answer:

rjkz [21]2 years ago

5 0

MA= output force/input force
MA= 100N/20N
MA= 50

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Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden

LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

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3 years ago

When magnesium ribbon is ignited, it gives off a white fume of magnesium oxide. What is this type of reaction?

Svetllana [295]

This reaction is called a composition reaction.

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3 years ago

What is the electron configuration of an element with atomic number 20?

VARVARA [1.3K]

1s2,2s2.2p6,3s2,3p6,3d4,4s2

8 0

2 years ago

Read 2 more answers

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

2 years ago

An electron is present at an energy level with an energy of 16.32 x 10−19 J. During a transition to another level, 5.4 x 10−19 J

kolbaska11 [484]

Answer:

D. The electron moved up to an energy level and has an energy of 21.72 x 10−19 J.

3 0

2 years ago