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V125BC
12 days ago
9

What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?

Chemistry
1 answer:
VMariaS [1K]12 days ago
5 0
<span>Answer: The answer varies based on the type of acid following "0.5440 M H...". For a monoprotic acid such as HCl, the calculation proceeds as follows: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH In the case of a diprotic acid like H2SO4, it works out to: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH For a triprotic acid such as H3PO4, it is: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH</span>
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The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?
castortr0y [923]

Greetings!

The result is:

The new volume is: 2L

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

P_{1}V_{1}=P_{2}V_{2}

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm

Let's put the values into the equation:

2.5L*6.0atm=7.5atm*V_{2}

2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L

Consequently, the new volume is: 2L

Wishing you a lovely day!

7 0
2 days ago
Acetone major species present when dissolved in water
Alekssandra [968]

The compound is acetone ( CH₃-CO-CH₃)


Explanation:


1) Acetone is represented as CH₃-CO-CH₃.


2) This is a molecule formed by covalent bonds.


3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.


In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.


This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).



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11 days ago
A hardware salesman measures the mass of a box containing
lorasvet [956]
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8 0
15 days ago
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Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added
alisha [964]

1) reacting hydrochloric acid with nickel:

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Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

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Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

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castortr0y [923]
Convert 55.0g Ca(OH)2 to moles.

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To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.

If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
4 0
6 days ago
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