All categories

3 years ago

Aluminum hydroxide + nitric acid ➡ aluminum nitrate + water

Chemistry

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

Chemical equation:

HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O

Explanation:

Chemical equation:

HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O

Balanced chemical equation:

3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O

Ionic equation:

3H⁺ + 3NO⁻₃(aq) + Al(OH)₃(s) → Al³⁺(aq) + 3NO₃⁻¹(aq) + 3H₂O(l)

Net ionic equation:

Al(OH)₃(s) + 3H⁺(aq) → Al³⁺(aq) + 3H₂O(l)

The NO⁻₃ are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

You might be interested in

How do you detemine the charge of an element based on jts valence electrons

mafiozo [28]

Look at the protons and electrons and rememeber metals want to get rid of electrons to be perfect. Nonmetals are wanting to take in. Metals are negatives in electrons. Opposite for nonmetals.

8 0

3 years ago

A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph

Mashutka [201]

We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.

5 0

3 years ago

A student is taking notes about the four steps in hurricane formation. Which of the following should be written as step number 1

Brrunno [24]

Tropical disturbance is the correct answer

3 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$