The process of converting nitrogen into nitrogen compounds by bacteria is called:

A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/s2, and a slug is the mass of an object that wil

Darya [45]

Remember the formula as per the second Law of Newton: F = m*a

And also remember that the weight is the force with which the mass is attracted by the planet (or satellite in the case of the moon).

With that information you can answer the questions:

a) Weight = F = m*a

m = 175 slugs = 175 lbm

i) Earth

a = 32.17 ft/s^2

Weight on Earth = 175 lbm * 32.17 ft / s^2 = 5,629.75 poundal

ii) Moon

a = [1/6] 32.17 ft/s^2

Weight on the Moon = [1/6]*5,629.75 poundal = 938.29 poundal

b) Force = 355 poundal
m = 25.0 slug

a in m/s^2 = ?

First calculate the force in ft/s^2

F = m*a => a = F/m = 355 poundal / 25.0 slug = 14.2 ft/s^2

Conversion:

14.2 ft / s^2 * [ 0.3048 m/ft] = 4.32816 m/s^2

Answer: 4.33 m/s^2

8 0

3 years ago

Choose all the answers that apply.

Anni [7]

Answer:

I believe it's the lowest portion of the atmosphere

3 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

I need help ASAP 10 points

jekas [21]

Answer:

the sun beams down on the pool and heats it up top to bottom the deeper the colder

4 0

3 years ago

What is the application of Complexometric titration?

PtichkaEL [24]

Explanation:

Complexometric titration (sometimes chelatometry) is a form of volumetric analysis in which the formation of a colored complex is used to indicate the end point of a titration. Complexometric titrations are particularly useful for the determination of a mixture of different metal ions in solution.

hope it will help u ✌️

7 0

3 years ago