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3 years ago

15

A solution of the weak acid HF and a solution of the strong acid HCl have the same pH. Which solution will require the most sodi

um hydroxide, NaOH, to neutralize

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

Explanation:

I think at HF is the answer because the pH value of weak acid are 4 and 5

please mark the brainliest if you are satisfied thank you very much

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A weather map is an example of a

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C. Model. It's a graphical model that displays one or more weather elements.

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2 years ago

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Write the balanced chemical equations for the decomposition of solid calcium hydroxide into solid calcium (ii) oxide (lime) and

Lesechka [4]

Calcium oxide (CaO) or lime in solid form can be prepared from the decomposition of calcium hydroxide {Ca $(OH)_{2}$ in to lime (CaO) and water ( $H_{2}O$ at high temperature. The reaction is an endothermic reaction. That is heat is absorbed in this reaction process. One mole of calcium hydroxide decomposed into one mole of calcium oxide and one mole of water. The balanced reaction can be shown as-CaC $O_{3}$ (solid) → CaO (solid) + $H_{2}O$ (liquid). The heat of the reaction is (+) 63.7 kJ/mole of CaO.

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3 years ago

Aluminum chloride can be formed from its elements:

saul85 [17]

<u>Answer:</u> The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

2 years ago

Irina-Kira [14]

Answer:

Double replacement

Explanation:

The given reaction is double replacement reaction.

CaCO₃ + 2HCl → CaCl₂ + H₂CO₃

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

while,

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

Decomposition:

It is the reaction in which one reactant is break down into two or more product.

AB → A + B

Single replacement:

It is the reaction in which one elements replace the other element in compound.

AB + C → AC + B

7 0

3 years ago

What is the molarity of a solution of 17.0 g of nh4br in enough h2o to make 158 ml of solution? answer in units of m?

vivado [14]

Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.

Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.

So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M

6 0

3 years ago