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3 years ago

What are the oxidation numbers for Zn(OH)4 ^2-

1 answer:

6 0

Zn(OH)4= ; -2 = -4 + ?, Zn = +2 , the -4 is the charge on the 4 OH- ions.

NO3- + 6 H2O + 8 e- -------> NH3 + 9 OH- 
( Zn + 4 OH- ----> Zh(OH)4= + 2e-) x 4 
7 OH- + NO3- + 4 Zn + 6 H2O -------> NH3 + 4 Zn(OH)4=

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Once the sun's energy reaches the earth's atmosphere through radiation, it is circulated within the atmosphere and oceans throug

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The sun's energy would be circulated through convection. This is because whenever you see the word "circulate", there is a pretty good chance it has something to do with convection.

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4 years ago

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8. Which is Earth's path around the sun?

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Answer:

earth's tilt

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3 years ago

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Can somebody please help me asap !!!

EastWind [94]

Answer:

Option A. 1191.49 K

Explanation:

Data obtained from the question include:

The equation for the reaction is given below:

4HCl + O2 —> 2Cl2 + 2H2O

Enthalpy (H) = +280 KJ/mol = +280000 J/mol

Entropy (S) = +235 J/Kmol

Temperature (T) =..?

The temperature at which the reaction will be feasible can be obtained as follow:

Change in entropy (ΔS) = change in enthalphy (ΔH)/T

(ΔS) = (ΔH)/T

235 = 280000/T

Cross multiply

235 x T = 280000

Divide both side by 235

T = 280000/235

T = 1191.49 K

Therefore, the temperature at which the reaction will be feasible is 1191.49 K

4 0

4 years ago

I need help ASAP please

Andrej [43]

525L

Explanation:

Given parameters:

Volume to Temperature ratio = 1.75

Temperature = 300k

Unknown:

Volume of the gas = ?

Solution:

We must carefully analyze and comprehend the detailed description of this experiment:

It was suggest that the gas could expand or shrink without changing the pressure of the gas inside.

This implies that the pressure in the vessel is constant

This similar to the postulate of the Charles's law: The volume of a fixed mass of gas is directly proportional to the absolute temperature if the pressure is constant.

Mathematically;

$\frac{V1}{T1} = \frac{V2}{T2}$

V and T are the volume and temperature

Now we can solve the given problem;

$\frac{V}{T}$ = 1.75

Since temperature is given as 300K

Input the variables:

$\frac{V}{300}$ = 1.75

V = 1.75 x 300 = 525L

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

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3 years ago

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most rem

son4ous [18]

Answer:

10 kg Al(OH)₃

Explanation:

There is some info missing. I think this is the original question.

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th-century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today.

In the first step, aluminum hydroxide reacts to form alumina (Al₂O₃) and water: 2 Al(OH)₃(s) → Al₂O₃(s) + 3H₂O(g). In the second step, alumina (Al₂O₃ and carbon react to form aluminum and carbon dioxide: 2Al₂O₃(s)+3C(s)→4Al(s)+3CO₂(g). Suppose the yield of the first step is 63% and the yield of the second step is 89%.

Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Let's consider the 2 steps in the synthesis of Al.

Step 1: 2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

Step 2: 2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)

In Step 2, the percent yield of Al is 89% and the real yield is 2.0 kg. The theoretical yield is:

2.0 kg (R) × (100 kg (T) / 89 kg (R)) = 2.2 kg = 2.2 × 10³ g

In Step 2, the mass of Al is 4 × 26.98 g = 107.9 g and the mass of Al₂O₃ is 2 × 101.96 g = 203.92g. The mass of Al₂O₃ that produced 2.2 × 10³ g of Al is:

2.2 × 10³ g Al × (203.92g Al₂O₃ / 107.9 g Al) = 4.2 × 10³ g Al₂O₃

In Step 1, the percent yield of Al₂O₃ is 63% and the real yield is 4.2 × 10³ g. The theoretical yield is:

4.2 × 10³ g (R) × (100 g (T)/ 63 g (R)) = 6.7 × 10³ g

In Step 1, the mass of Al₂O₃ is 101.96 g and the mass of Al(OH)₃ is 2 × 78.00 g = 156.0 g. The mass of Al(OH)₃ that produced 6.7 × 10³ g of Al₂O₃ is:

6.7 × 10³ g Al₂O₃ × (156.0 g Al(OH)₃ / 101.96 g Al₂O₃) = 1.0 × 10⁴ g Al(OH)₃ = 10 kg Al(OH)₃

7 0

4 years ago