All categories

3 years ago

C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O

Chemistry

1 answer:

Vlad [161]3 years ago

4 0

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

You might be interested in

What is the law of reciprocal proportion

irina [24]

If two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same as or a simple multiple of the ratio of the masses in which they combine with each other.

7 0

2 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

Calculate the change in energy when 75.0 grams of water drops from 31.0C to 21.6.

zysi [14]

Answer: Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

5 0

3 years ago

Read 2 more answers

What happens to the cell membrane during exocytosis?

Cloud [144]

Answer:

Endocytosis and Exocytosis: Differences and Similarities

ARTICLE Apr 28, 2020

by Nicole Gleichmann

Endocytosis and Exocytosis: Differences and Similarities

Endocytosis and exocytosis are the processes by which cells move materials into or out of the cell that are too large to directly pass through the lipid bilayer of the cell membrane. Large molecules, microorganisms and waste products are some of the substances moved through the cell membrane via exocytosis and endocytosis.

Why is bulk transport important for cells?

Cell membranes are semi-permeable, meaning they allow certain small molecules and ions to passively diffuse through them. Other small molecules are able to make their way into or out of the cell through carrier proteins or channels.

But there are materials that are too large to pass through the cell membrane using these methods. There are times when a cell will need to engulf a bacterium or release a hormone. It is during these instances that bulk transport mechanisms are needed.

Endocytosis and exocytosis are the bulk transport mechanisms used in eukaryotes. As these transport processes require energy, they are known as active transport processes.

Vesicle function in endocytosis and exocytosis

During bulk transport, larger substances or large packages of small molecules are transported through the cell membrane, also known as the plasma membrane, by way of vesicles – think of vesicles as little membrane sacs that can fuse with the cell membrane.

Cell membranes are comprised of a lipid bilayer. The walls of vesicles are also made up of a lipid bilayer, which is why they are capable of fusing with the cell membrane. This fusion between vesicles and the plasma membrane facilitates bulk transport both into and out of the cell.

What is endocytosis? Endocytosis definition and purposes

Endocytosis is the process by which cells take in substances from outside of the cell by engulfing them in a vesicle. These can include things like nutrients to support the cell or pathogens that immune cells engulf and destroy.

Endocytosis occurs when a portion of the cell membrane folds in on itself, encircling extracellular fluid and various molecules or microorganisms. The resulting vesicle breaks off and is transported within the cell.

8 0

2 years ago

Consider an ice cube and a hot radiator. Which has the higher thermal energy?

Law Incorporation [45]

The answer to your question is the radiator

8 0

2 years ago