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podryga [215]
1 year ago
6

6.

Chemistry
1 answer:
RUDIKE [14]1 year ago
3 0

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

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In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R =  0.0821 atm-L/mol-K

So,

V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L

So, the volume of the container is 59.112 L

6 0
1 year ago
Which of the reagents listed below would efficiently accomplish the transformation of ethyl-3-pentenoate into 3-penten-1-ol?
Mandarinka [93]
Reactions of Ethyl-3-pentenoate with all given reagents are given below.

Reaction with H₂ / Pd:
                                     
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.

Reaction with NaBH₄:
                                   Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.

Reaction with LiAlH₄:
                                  Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.

Result:
           The correct answer is Option-A (Highlighted RED below).

7 0
2 years ago
If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
The first ionization energy, e, of a potassium atom is 0.696 aj. what is the wavelength of light, in nm, that is just sufficient
elena55 [62]
The formula to be used for this problem is as follows:

E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J

0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
6 0
1 year ago
Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg
natita [175]

Answer:

a.

N = +2

O = -2

b.

Na = +1

O = -2

N = +5

c.

O = -2

Al = +3

P = +5

d.

O = -2

Ca = +2

C = +4

e

O = -2

Mn = +7

K = +1

f.

O = -2

K = +1

Cr = +6

g.

O = -2

Cl = +7

Cr = +1

h.

O = -2

H = +1

Cr = +5

i.

O = -2

H = +1

Cr = +3

k.

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

d. carbonat de calciu

Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0
1 year ago
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