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3 years ago

In the soap testing experiment, what is the purpose of adding some MgCl2 to the soap solution at the end?

Chemistry

1 answer:

lakkis [162]3 years ago

4 0

Explanation:

The water is hard due to the presence of ions of Mg²⁺. Now in hard water soaps are ineffective . Since in hard water Mg²⁺ ion forms precipitate , which concludes less number of soap molecules are present in the solution and less amount of froth which disables the cleansing property of the soap .

Now the experiment , the very last step is the hard water test where some amount of MgCl₂ is added ,

Now this magnesium salt act as a source of Mg²⁺ ion and now the soap action can be determined , whether it is able to form froth or not .

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A single alkyl bromide reactant theoretically yields either of the given products, depending on the reaction conditions. Draw th

Ierofanga [76]

Answer:

$H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$

Explanation:

The steps involved in predicting the structure of the alkyl bromide compound are outlined below.

1) An examination of the product shows that the product could only be formed by a substitution reaction.

2) The structure of the alkyl bromide compound can be then predicted by replacing the methoxide group in the product after the substitution of bromine atom. This is because the methoxide ion acts as a strong nucleophile.

Therefore, by consideration the reaction mechanisms of reactions 1 and 2, it can be predicted that the structure of the alkyl bromide compound is $H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$ . A pictorial diagram of the alkyl bromide compound is also attached.

8 0

3 years ago

Read 2 more answers

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre

Marina CMI [18]

Answer:

$\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}$

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

$n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}$

2. Moles of NH₃

$n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}$

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

$c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}$

(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

3 0

3 years ago

Calculate the standard free energy for the following reaction at 25Â°C.

sladkih [1.3K]

I dont think with this much amount of information we can solve this...unless its an reversible reaction in that case free energy =0

3 0

3 years ago

What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed

monitta

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH = 1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

= 32.042 g/mol

Molar Mass of CO₂ = 1(12.01 g/mol) + 2(16.00 g/mol)

= 44.01 g/mol

Mass of only one reactant i.e. CH₃OH is given so it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

2 mol CO₂ / 2 mol CH₃OH

Convert moles of CO₂ to grams of CO₂ utilizing molar mass of CO₂ as:

44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ] [1 mol CH₃OH / 32.042 g CH₃OH ] [2 mol CO₂ / 2 mol CH₃OH] [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

= 14.28 g CO₂

Theoretical yield = 14.28 g CO₂

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

= 0.707 x 100%

= 70.7 %

Hence option A 70.7% yield is the correct answer.

8 0

4 years ago

Why do we call m commerce as handy device?

Jobisdone [24]

Answer:

m - commerce is the buying and selling of goods and service through wireless handheld devices such as smartphones and tablets as a form of commerce, m - commerce enables users to access online shopping plat forms without needing to use a desktop computer

3 0

3 years ago