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3 years ago

According to your results, rank the household products that you have used according to the strength of acidity?

1 answer:

4 0

strongest being phenolic acid with the highest acidity for cleaning very set in stains next it is malt vinegar good for toilets, cane vinegar good for floors and skirting board and white wine vinegar good for mixing with citric acid to clean and as already previously stated citric acid

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Inclined planes reduce the amount of effort needed to move an object, but increases

QveST [7]

Inclined planes reduce the amount of effort needed to move an object, but increases the length of the ramp.

Explanation:

Mechanical advantage is the measure of amount of effort needed to move an object. The mechanical advantage can be calculated as the ratio of length of ramp to the height of ramp for an inclined plane.

As it is known that an object can be easily moved on an inclined plane than on a vertical plane, this is because, the inclined plane provides greater output force. But in that case, the effort required will be reduced with the cost of increasing the distance of the movement of object.

In other terms , the ramp's length of inclined planes has to get increased in order to reduce the amount of effort needed to move an object. This is because as the mechanical advantage has length of the ramp in the numerator, with the increase in numerator value or length value the mechanical advantage will also increase.

8 0

3 years ago

Distinguish between zinc nitrate and lead nitrate

Kobotan [32]

Answer:

Zinc nitrate gives white ppt. which dissolves in excess ammonium hydroxide and produce a colorless solution whereas lead nitrate gives a chalky white ppt. of lead hydroxide which doesnot dissolve.

Explanation:

Hope this helps :)

7 0

3 years ago

The iron content of foods can be determined by dissolving them in acid (forming Fe3+), reducing the iron(III) to iron(II), and t

____ [38]

Answer:

Oxidation half-reaction:

Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻

Reduction half-reaction:

Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)

Explanation:

The reaction that takes place is:

Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)

The oxidation half-reaction is:

Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The reduction half-reaction is:

Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

4 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Why is the atmosphere of Earth the outermost layer of the planet?

nasty-shy [4]

Option C but i am not sure

3 0

2 years ago

Read 2 more answers