What is the density (g/L) of CHCl3 vapor at 1.00 atm and 298 K?

The values used in the scale of pH and pOH are derived from a system designed by ______. Gordonsen Sorenson Curie Dalton

yaroslaw [1]

Sorenson

Explanation:

The values used in the scale of pH and pOH are derived from a system designed by Sorenson. Søren Peter Lauritz Sørensen, a Danish chemist introduced the system of pH and pOH for describing the alkalinity and acidity of substances.

The pH and pOH scale is logarithmic scale that ranks the acidity and bascity of compounds.
pH is the negative logarithm of the concentration of hydrogen/hydroxonium ions in solution i.e

pH = -log₁₀{H⁺]

pOH is the negative log of the concentration of the hydroxyl ions in a solution i.e

pOH = -log₁₀{OH⁻]

Learn more:

calculating pH: brainly.com/question/12985875

pH scale: brainly.com/question/11063271

#learnwithBrainly

3 0

2 years ago

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, <u>we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition</u>. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

<u>Where</u>:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

7 0

2 years ago

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of

allochka39001 [22]

Answer:

The answer to your question is C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

12 g of C ------------------ 1 mol

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

1 g of H -------------------- 1 mol of H

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

16 g of O ---------------- 1 mol

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O

8 0

2 years ago

Describe a single measurement that you could make that shows that a kettle is less than 100% efficient (providing evidence that

alexgriva [62]

Answer:

As you haven't explained what measurements you took before solving this problem, I will explain the general procedure to evaluate the efficiency of a kettle. I hope it helps you. I´ll send an attachement file with the full answer, since I couldn't write it here.

I assume that the material that is going to be heated in the kettle is water.

1- You have to boil water in it and take the time it takes to its boiling point (in seconds).

2- You have to evaluate the amount of energy the water absorbed Q with the efficiency formula which I explain in the attachement file.

3- Divide Q by the time it took to bring the water to boiling so you can have the power it consumed.

4- You divide the last value you obtained by the Kettles's power rating.

5- Multiply the last value by 100 to obtain a percentage value of efficiency.

Explanation:

Efficiency is the ration of a machine's useful work, in this case how much energy the water absorbed to get to its boiling point divided by the time it took to get to this point, and the total energy expended, in this case the kettles's power rating.

7 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago