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2 years ago

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Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to

electroplate the zinc to a thickness of 0.025 mm. It is found that a current of 20 A produces a zinc coating of sufficient quality for galvanized iron. Determine the time required to produce the desired coating, assuming 100 % efficiency.

1 answer:

Alexandra [31]2 years ago

8 0

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode $(m_{ch})$ is given as :

$\frac{m_{ch}}{M} ZF = It$

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; $(m_{ch})$ = $(Ad) \rho$ ; replacing that into above equation; we have:

$\frac{(Ad) \rho}{M} ZF = It$ ---- equation (1)

where;

A = area

d = thickness

$\rho$ = density

From the above equation (1); The time required for coating can be calculated as;

$[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t$

$t = \frac{2100}{20}$

= 105 s

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Combustion of 8.652 g of a compound containing c, h, o, and n yields 11.088 g of coz, 3.780 g of h2o, and 3.864 g of no2. How ma

Katyanochek1 [597]

Answer:- C = 3.024 g, H = 0.42 g, N = 1.176 g and O = 4.032 g

Solution:- The compound contains C, H, N and O. On combustion, all the carbon is converted to carbon dioxide, All hydrogen is converted to water and all the nitrogen is converted to nitrogen dioxide.

From the grams of all these, we could calculate their moles and then using mol ratio of these products and the number of moles of C, H and N present in them, we calculate the moles of C, H and N respectively. Further, these moles are converted to grams. On subtracting the sum of grams of C, H and N from the mass of the sample, the mass of oxygen is calculated.

The calculations are as follows:

Calculations for the grams of C:-

$11.088gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})(\frac{12gC}{1molC})$

= 3.024 g C

Calculations for the grams of H:-

$3.780gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})(\frac{1gH}{1molH})$

= 0.42 g H

Calculations for the grams of N:-

$3.864gNO_2(\frac{1molNO_2}{46gNO_2})(\frac{1molN}{1molNO_2})(\frac{14gN}{1molN})$

= 1.176 g N

Mass of the compound is given as 8.652 g. Now we could calculate the grams of oxygen as:

mass of oxygen = 8.652 - (3.024 + 0.42 + 1.176)

= 8.652 - 4.62

= 4.032 g

So, 8.652 grams of the compound contains 3.024 g of C, 0.42 g of H, 1.176 g of N and 4.032 g of O.

7 0

1 year ago

How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt? Give your answer in kg/day.

ser-zykov [4K]

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0 $m^{3}/s$

1 $m^{3}/s$ = 1000L

Multiply by 1000 to convert to L/s

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

= 2,592,000,000L/day

if the river contains 50mg of salt in 1L of solution

lets find how many mg of salt (X) is contained in 2,592,000,000L/day

X= $\frac{ 2,592,000,000*50}{1}$

X= 129,600,000,000 mg/day

convert this value to kg/day by multiply by $10^{-6}$

X= 129,600kg/day

7 0

2 years ago

The molarity of a solution of 5.0 g of kcl in 100. ml of solution is ________.

sdas [7]

Molarity is defined as the number of moles of solute in 1 L of solution
mass fo KCl in the solution is - 5.0 g
number of moles of KCl - 5.0 g/ 74.5 g/mol = 0.067 mol
number of moles of KCl in 100 mL - 0.067 mol
therefore number of KCl moles in 1 L - 0.067 / 100 mL x 1000 mL = 0.67 M
molarity of KCl is 0.67 M

7 0

1 year ago

Read 2 more answers

Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?

Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

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1 year ago

Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp

sergiy2304 [10]

Explanation:

The given data is as follows.

$t_{m}$ = 30.0 sec, $t_{r1}$ = 5 min = $5 \times 60 sec$ = 300 sec

$t_{r2}$ = 12.0 min = $12 \times 60 sec$ = 720 sec

Formula for adjusted retention time is as follows.

$t'_{r} = t_{r} - t_{m}$

= 300 sec - 30.0 sec

= 270 sec

$t'_{r2}$ = 720 sec - 30 sec

= 690 sec

Formula for relative retention ( $\alpha$ ) is as follows.

$\alpha = \frac{t'_{r2}}{t'_{r1}}$

= $\frac{690 sec}{270 sec}$

= 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0

2 years ago