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2 years ago

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5 0

Ammonium hydroxide is considered a weak base because it produces few hydroxide ions in aqueous solution. (Option a)

<h3>Strength of electrolytes</h3>

Strong electrolytes: are completely ionized in aqueous solution.
Weak electrolytes: are partially ionized in aqueous solution.

Let's consider the ionization of ammonium hydroxide, which is a weak base.

NH₄OH ⇄ NH₄⁺ + OH⁻

Why is it considered a weak base?

a. it produces few hydroxide ions in aqueous solution. Yes, since it dissociates partially.

b. it produces few hydrogen ions in aqueous solution. No, since it does not produce hydrogen ions.

c. it does not conduct electricity in aqueous solution. No, since it conducts electricity, although poorly.

d. it will not take part in a neutralization reaction. No, since it will be neutralized by acids.

Ammonium hydroxide is considered a weak base because it produces few hydroxide ions in aqueous solution.

Learn more about weak bases here: brainly.com/question/322250

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Which of the solutions have greatest osmotic pressure 30% sucrose 60% sucrose or 30% magnesium sulfate?

timofeeve [1]

Answer:Osmotic pressure is the minimum amount of pressure a solution must exert in order to prevent from crossing a barrier by osmosis. Solute molecules have difficulty crossing semipermeable membranes, so the more solutes that are in a solution, the higher the osmotic pressure will be. Between 30% sucrose and 60% sucrose, 60% sucrose will have a greater osmotic pressure than 30% because it has a higher percentage of solutes. However, since sucrose has a higher potential to cross semipermeable membranes and is more absorbable than magnesium sulfate, magnesium sulfate would have a higher osmotic pressure than 60% sucrose even though 60% sucrose has higher molecules.

Explanation:

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3 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

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$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

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$\textsf{ \large{ \underline{Now substituting the required values}}}$

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$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

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Henceforth, the change in boiling point is 0.00735°C.

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1 year ago

Importance of pH value in daily life

olganol [36]

Plants require pH to thrive which in turn gives us food.

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3 years ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)