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3 years ago

12

soaps do not work well in hard waters. But they can still be effected it large quantities of soap are added to water. Explain wh

y.

2 answers:

Nookie1986 [14]3 years ago

6 0

They can still be effected it large quantities of soap are added to water

because more soap eliminates the effect the minerals in hard water has on

its cleaning capacity.

Hard water contains minerals such as

Calcium

Magnesium

Iron etc.

They reduce the cleaning capacity of the soap. This is why soft water is

preferably used as it doesn't contain these minerals.

When more soap is added, the effect of the minerals reduces thereby

bringing about very little or no effect on the cleaning capacity of the soap.

Read more about Soap here brainly.com/question/1473301

Yakvenalex [24]3 years ago

5 0

Explanation:

Ordinary water does not remove dirt from clothes or skin because the dirt present is oily or greasy in nature. Soaps are one of the most commonly used cleansing agents and are capable of reacting with water to remove dirt. They are either animal origin or plant in origin.

Chemically, they are water soluble sodium or potassium salts of higher fatty acids such as stearic acid, palmitic acid or oleic acid. Fatty acids are found in fats and oils. When fat or oil is treated with alkali such as caustic soda or caustic potash soap and glycerin are produced.

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You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

2) Which of the following is a synthesis reaction?

Artemon [7]

Answer:

(4) CuO + H₂ → Cu + H₂O.

Explanation:

<em>(1) Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag.</em>

It is considered as a single replacement reaction that Cu replaces Ag in its salt AgNO₃.

<em>(2) 2Cu + O ₂ → 2CuO.</em>

It is a combustion reaction that Cu reacts with oxygen results CuO.

<em>(3) CuCO₃ → CuO + CO₂.</em>

It is a decomposition reaction that CuCO₃ is decomposed to CuO and CO₂.

<em>(4) CuO + H₂ → Cu + H₂O.</em>

It is considered as a synthesis reaction that two reactants CuO and hydrogen are reacted to produce Cu and H₂O.

<em>So, the right choice is: (4) CuO + H₂ → Cu + H₂O.</em>

<em></em>

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4 years ago

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Define Titanium and what it is used for

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Answer:

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You are given a clear solution of KNO3. Using 3-4 sentences explain how you would determine if the solution is unsaturated,satur

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3 years ago