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Alexus [3.1K]
3 years ago
12

Suppose a thin sheet of zinc containing 0.2 mol of the metal is completely converted in air to zinc oxide (zno) in one month. ho

w would you express the rate of conversion of the zinc?
Chemistry
2 answers:
hram777 [196]3 years ago
8 0
Molecular mass of  Zn = 65.38  
<span>1 mol = 65 g </span>
<span>0.2 mol of Zinc = 13.076 g of Zinc </span>

<span>13.076 g of Zinc is completely converted in air to ZnO </span><span>in one month
= 13.076 g / 30 days = 0.435 g per day 
</span>
<span>The rate of conversion of the zinc = 0.44 g / day</span>
MissTica3 years ago
7 0

Answer:

r_{Zn}=-kC_{Zn}^2C_{O_2}

Explanation:

Hello,

In this case, the mentioned reaction should be:

2Zn(s)+O_2(g)\rightarrow 2ZnO

In such a way, considering an irreversible reaction and an elemental reaction, the power of each compound's concentration equals the stoichiometric coefficient in its reaction, thus:

r_{Zn}=-kC_{Zn}^2C_{O_2}

It turns out negative since it is a consumption upon zinc.

Best regards.

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11111nata11111 [884]

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V

4 0
2 years ago
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

3 0
3 years ago
Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Draw the st
pentagon [3]

Answer:

Explanation:

Structure of the 2,2,4,4-tetramethyl-3-pentanone is give in the attachment

In 2,2,4,4-tetramethyl-3-pentanone, no alpha hydrogen is present, therefore, enol form is not possible and hence, exist only in keto form.

Explanation for existence of cyclohexa-2,4-diene-1-one only in enol form:

keto form of cyclohexa-2,4-diene-1-one not aromatic and hence less stable.

Whereas enol form it is aromatic which makes it highly stable. that's why cyclohexa-2,4-diene-1-one exists only in enol form.

8 0
3 years ago
How did Mendeleev arrange the known elements?
Agata [3.3K]
In order of relative atomic mass.
4 0
3 years ago
Read 2 more answers
A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina
NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

3 0
3 years ago
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