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3 years ago

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An electrolytic cell can be used to plate gold onto other metal surfaces. The plating reaction is: Au (aq) e- --> Au(s) Notic

e from the reaction that 1 mol e- plates out 1 mol Au(s). Use this stoichiometric relationship to determine how much time is required with an electrical current of 0.200 amp to plate out 0.400 g Au. The amp is a unit of electrical current equivalent to 1 C / s. (Hint: Recall that the charge of an electron is 1.60 x 10-19 C.)

1 answer:

Dmitriy789 [7]3 years ago

4 0

Answer:

Explanation:

Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time

Q = I × t = n × F

Where

Q is charge

I is current = 0.2A

t is time = ?

n is number of mole

F is Faraday constant = 96500C/s

Mass of Au = 0.4 g = 0.0004kg

Molar mass of Au = 197 g/mol

Malar mass = mass / mole

n = 0.4 / 197

n = 2.03× 10^-3 mol

n = 0.00203 mol

t = n × F / I

t = 0.00203 × 96500 / 0.2

t = 979.7 seconds

t = 16.32 minutes.

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In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m

lara31 [8.8K]

Answer:

$F=3.61\times 10^{-47}\ N$

Explanation:

Mass of a proton, $m_p=1.67\times 10^{-27}\ kg$

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The distance between the electron and the proton is, $r=5.3\times 10^{-11}\ m$

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

$F=G\dfrac{m_em_p}{r^2}$

Where G is the universal Gravitational constant

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So, the force between the electron and proton is $3.61\times 10^{-47}\ N$ .

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2 years ago

Which branch of science involves the study of matter and energy

Olin [163]

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3 years ago

Read 2 more answers

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

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2 years ago

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Answer:

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3 years ago

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