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Tasya [4]
3 years ago
10

Chromium metal is plated onto a car bumper at a current of 10.0 amperes for 100 seconds. The plating bath contains chromium in t

he form of CrO42− ions. How many grams of Cr are deposited in this process? (The atomic weight of Cr is 51.996 amu.)
Chemistry
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

0.54g of Cr

Explanation:

Current (I) = 10A

Time (t) = 100s

Molecular mass of Cr = 51.996 amu

Faraday's first law of electrolysis states that

The mass of the substance (m) of a given substance deposited at an electrode is directly proportional to the quantity of electricity or charge (Q) passed

m = nQ

M = mass of the substance

n = electrochemical constant

Q = charge passed through it

Q = IT

Q = (10 * 100) = 1000C

1 moles = molarmass = Faraday's constant (96500C)

Molar mass = Faraday's constant (96500C)

51.996 g = 96500C

How many grams will be liberated with 1000C

51.996g = 96500C

Xg = 1000C

X = (1000 * 51.996) / 96500

X = 51996 / 96500

X = 0.5388g = 0.54 g of Cr will be deposited

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1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Hospital patients are administered oxygen from an pressurized
VashaNatasha [74]

Answer:

32000atm

Explanation:

Using Boyle's law equation;

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (

V2 = final volume (L)

According to the question below:

P1 = 160.0 atm

P2 = 3.0 atm

V1 = 600L

V2 = ?

Using P1V1 = P2V2

160 × 600 = 3 × V2

96000 = 3V2

V2 = 96000/3

V2 = 32000atm

5 0
3 years ago
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