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3 years ago

What is the mass of 11.5 liters of CL2 gas at STP?

1 answer:

6 0

At STP, the volume of Cl2 is 22.4 L/mol.

11.5 L Cl2 * (1 mol Cl2/ 22.4 L Cl2)= 115/224 mol Cl2.

115/224 mol Cl2* (70.90 g Cl2/ 1 mol Cl2)= 36.4 g Cl2.

The final answer is 36.4 g Cl2.

Hope this helps~

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A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC . At this

Greeley [361]

Answer:

The mol fraction of cyclohexane in the liquid phase is 0.368

Explanation:

Step 1: Data given

Mass of cyclohexane = 25.0 grams

Mass of 2-methylpentane = 44.0 grams

Temperature = 35.0 °C

The pressure of cyclohexane = 150 torr

The pressure of 2-methylpentane = 313 torr

The pressure we only need for the mole fraction in gas phase.

Step 2: Calculate moles of cyclohexane

Moles cyclohexane = mass cyclohexane / molar mass

Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane

Step 3: Calculate moles of 2-methylpentane

Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane

Step 4: Calculate mole fraction of cyclohexane in the liquid phase

Mole fraction of C6H12:

0.298 / (0.298 + 0.512) = 0.368

The mol fraction of cyclohexane in the liquid phase is 0.368

5 0

4 years ago

1. For CCl4, upload a photo including the following: (1) work arriving at total # valence electrons

forsale [732]

Work arriving at total valence electrons are 24

The valence electron is an electron in the outer shell associated with an atom, and that can participate in the formation of a chemical bond

For a main-group element, a valence electron can exist only in the outermost electron shell; for a transition metal, a valence electron can also be in an inner shell.

Lewis structures, also known as Lewis dot formulas, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDS), are diagrams that show the bonding between atoms of a molecule, as well as the lone pairs of electrons that may exist in the molecule.

Calculating the work arrival at total valence electrons

We know 1 bond contains

2 valence electrons

Here CCl4 contains 4 bonds that is 8 valence electrons

Therefore ,

Total valence electrons are 32

Now valence electrons remain after bond formation =

32 - 8

=24 valence electrons

A molecular model is a physical model of an atomistic system that represents molecules and their processes

Molecular image and lewis hand drawn structure is as shown below

Learn more about lewis structure here

brainly.com/question/6224949

#SPJ10

3 0

2 years ago

a 20ml sample of hcl was titrated with the 0.0220 M Naoh. to reach the endpoint required 23.72 mL of the NaOh. Calculate the mol

Y_Kistochka [10]

Answer:

Molarity of HCl=0.026092M

Explanation:

The equation for the reaction is;

HCl + NaOH ⇒ NaCl + H2O

Using the formular, $\frac{C_{A}V_{A}}{C_{B}V_{B} }=\frac{nA}{nB}$ ..........equ1

where $C_{A}$ is the concentration of Acid,

$V_{A}$ is volume of acid

$C_{B}$ is concentration of the base

$V_{B}$ is volume of the base

nA is the number of moles of Acid

nB is number of moles of base

nA = 1, nB=1 , $V_{A}=20ml$ , $C_{B}=0.022M$ , $V_{B}=23.72mL$

we will input these values into equation1 to solve for $C_{A}$

$\frac{C_{A}*20}{0.022*23.72}=\frac{1}{1}$

$C_{A}*20=0.022*23.72$

$C_{A}=0.026092M$

7 0

3 years ago

How to 45 m/s converted to km/hr

DENIUS [597]

45 m/s * 3.6 = 162 km/h. Multiply with 3.6 to go to km:h, divide by 3.6 to go to m/s

5 0

3 years ago

Calculate the concentrations of h2so3, hso−3, so2−3, h3o+ and oh− in 0.025 m h2so3.

sammy [17]

We will use this two reaction equation:

H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2

HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

H2SO3 + H2O ↔ H3O+ + HSO3-

initial 0.025 0 0

change -X +X +X

Equ (0.025-X) X X

Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X

∴[H3O+] = 0.0127 M

and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

= 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

= 8.66 x 10^-13 M

- by using the ICE table for the second equation:

HSO3- + H2O ↔ H3O+ + SO3 2-

initial 0.0127 0.0127 0

change -X +X +X

Equ (0.0127-X) (0.0127+X) X

when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X

∴[SO32-] = 6.3 x 10^-8

3 0

4 years ago