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BigorU [14]
3 years ago
5

_HNO3 + _Al(OH)3

Chemistry
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

Option D. 3, 1, 3, 1

Explanation:

From the question given above,

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

The equation can be balance as follow:

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:

3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:

3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃

Now, the equation is balanced.

Thus, the coefficients are 3, 1, 3, 1

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When electrolyzing copper (ll) chloride, what reaction takes place at the anode? What reaction takes place at the cathode?
klemol [59]

Answer:

Copper ions are reduced into copper atoms.

Cu²⁺₍aq₎ + 2e⁻ →  Cu₍s₎

Explanation:

During electrolysis,  the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.

At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)

Thus the half equation is as follows:

Cu²⁺₍aq₎ + 2e⁻ →  Cu₍s₎

5 0
3 years ago
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
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