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3 years ago

What is the purpose of the technology design process?

1 answer:

4 0

Steps of the technological design process include: identify a problem, research the problem, generate possible solutions, select the best solution, create a model, test the model, refine and retest the model as needed, and communicate the final solution.

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The colligative molality of an unknown aqueous solution is 1.56 m.

yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- Vapor pressure lowering

ΔP = P° . Xm . i

- Boiling point elevation

ΔT = Kb . m . i

- Freezing point depressión

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0

3 years ago

Steve and Meg are doing an experiment where they need to make 37 grams of Fe2O3. How many grams of Fe would they need ?

Anna11 [10]

Answer:

25.9g

Explanation:

Molecular mass of Fe₂O₃ = (56 * 2) + (16 * 3)

= 112 + 48 = 160

Let's follow the unitary method now

In 160g Fe₂O₃ , there is 112 g Fe

In 1 g Fe₂O₃ , there is 112 / 160 g Fe

In 37 Fe₂O₃ , we need 112 * 37 / 160

= 25.9 g Fe

5 0

2 years ago

PLEASE HELP! IM TRYNA GET A GOOD GRADE! I ONLY HAVE 6 MINS LEFFTT

Arte-miy333 [17]

Answer:

<h2>answer - b molarity </h2>

Explanation:

<h2>please like me</h2><h2>I hope answer correct</h2>

6 0

2 years ago

________ is required to do work.

Leona [35]

Hello

Force is Required to work

Hope this helps

5 0

3 years ago

Arrange the gases from most dense to least dense at a given temperature and pressure.

aleksandr82 [10.1K]

Answer:

Arrange them in order of greatest to least Molar Mass to get most dense to least dense at a given temperature and pressure.

Explanation:

"Given Temperature and Pressure" just tells you that those two variables are being held constant. The Ideal Gas Law (PV = nRT) shows that n (the # of moles) and V (the volume) are variable still. To keep pressure constant, we need to increase the volume so that the force per area is the same which would unfortunately lower the density. The only way for a gas to have the greatest density would be to have the least average kinetic energy to keep the pressure constant at a smaller volume. This is because kinetic energy is equal to KE = 1/2 m v^2, where m is mass in kilograms and v is the velocity. IT IS CRUCIAL TO NOTE THAT ACCORDING TO KINETIC MOLECULAR THEORY, ALL GASES HAVE THE SAME AVG. KINETIC ENERGY WHEN AT THE SAME TEMPERATURE. So since the avg. KE is also constant across these answers due to the temp. being constant, our only way to decrease the velocity (the rate at which the particles are moving per second) and in turn the number of collisions (to repeat, a smaller number of collisions is desired because they are equal to the pressure that must maintained at a constant value at the smallest volume possible to achieve highest density possible) is to increase the molar mass. This is also pretty intutive for the other part of the issue: since Density is a broken heart (m/v), a higher molar mass would ensure a smaller pressure (due to decreased velocity) and smaller volume (due to constant pressure). Okay, now order the gases heaviest to lightest in terms of molar mass and you got your answer. Please gimme brainliest if this helped bc it took 15 minutes to type out. Hope this helped you! :D

5 0

2 years ago