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2 years ago

What portion of your salt solution in ice water system is: a. A mixture? b. A solution?

1 answer:

8 0

The salt and water are a homogeneous mixture but when salt dissolves in the water system is called a solution of salt and water.

<h3>What is a mixture? </h3>

A mixture is defined as the combination of two or more substances that are not chemically bonded together.

There are two types of mixture which include:

Homogeneous (uniform composition) and

Heterogeneous mixtures

When salt is added to the ice water system, it lowers the freezing point of the ice water thereby forming a homogenous mixture of water and salt.

The dissolution of salt in ice water leads to the formation of salt and water solution.

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A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0

3 years ago

Read 2 more answers

Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g

Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

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8 0

2 years ago

calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

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4 0

1 year ago

how does the percentage by mass of the solute describe the concentration of an aqueous solution os potassium sulfate

kirza4 [7]

Answer:

Explanation:

In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:

mass percentage = (mass of solute/total mass of solution) x 100%

For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.

3 0

2 years ago

What is the best place to find detailed information about a hazardous chemical used in your work area

azamat

Answer:

MSDS -- Material Safety Data sheet

Explanation:

The full form of MSDS is Material Safety Data sheet. A Material Safety Data Sheet is a document which contains the information related to the potential hazards (fire or reactivity and environmental and health ) and includes how to work or use the chemical product safely.

It is regarded as an essential starting point for development of the complete health as well as safety program. MSDS also contains other information such as the use of the chemical, its storage and also handling and the emergency procedures that are related to the hazards of the material.

3 0

3 years ago