Identify the spectator ions in the reaction that occurs between aqueous solutions of potassium hydroxide and hydrochloric acid

The balanced reaction between aqueous nitric acid and aqueous strontium hydroxide is ________.

Kaylis [27]

d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H 4H
8O 8O
2N 2N
1Sr 1Sr

3 0

3 years ago

What type of reaction is : CaCo3+ HCl = CaCl2 + CO2 +H2O

Pavlova-9 [17]

Answer:

Double decomposition reaction

Explanation:

When hydrochloric acid reacts with any carbonates/hydrogen carbonates the products formed are metal chloride , water and carbon dioxide.Since HCl decomposes salts of weaker acids.

So the equation of the reaction between calcium carbonate and HCl is:

CaCO3+2HCl =CaCl2+H2O+CO2

This reaction is also known as double decomposition reaction.

7 0

3 years ago

Read 2 more answers

25 °C = ________ K which one 298 0 33 197

Paraphin [41]

Answer:

298

hope this helps

have a good day :)

Explanation:

5 0

3 years ago

Read 2 more answers

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago

A stock solution of FeCl2 is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M soluti

Harman [31]

Answer:

31.5mL

Explanation:

The following were obtained from the question:

C1 (concentration of stock solution) = 2M

V1 (volume of stock solution) =.?

C2 (concentration of diluted solution) = 0.630M

V2 (volume of diluted solution) = 100mL

Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

2 x V1 = 0.630 x 100

Divide both side by 2

V1 = (0.630 x 100) /2

V1 = 31.5mL

Therefore, 31.5mL of 2M solution of FeCl2 required

7 0

3 years ago

Read 2 more answers

Identify the spectator ions in the reaction that occurs between aqueous solutions of potassium hydroxide and hydrochloric acid￼

Identify the spectator ions in the reaction that occurs between aqueous solutions of potassium hydroxide and hydrochloric acid