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2 years ago

The airplanes parts and functions

Chemistry

1 answer:

SpyIntel [72]2 years ago

6 0

Answer: The main sections of an airplane include:

fuselage: The plane’s body, or fuselage, holds the aircraft together, with pilots sitting at the front of the fuselage, passengers and cargo in the back.

Wings: An aircraft’s wings are critical to flight through the production of lift, but they have many parts of the wing to control this lift amount and direction.

Cockpit: The cockpit is the area at the front of the fuselage from which a pilot operates the plane

Engine: The engine(s), or powerplant, of an aircraft creates thrust needed for the plane to fly.

Propeller: An aircraft’s propeller(s) are airfoils, similar to a wing, installed vertically to create thrust to drive the plane forward.

Tail assembly: An aircraft’s tail is mainly used for stability, as well as creating lift in combination with the wings. It’s comprised of several parts.

Landing gear: Landing gear is located under the belly of the plane consisting of a wheel and strut to soften impact with the ground and may be retractable into the fuselage.

Explanation:

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You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol

Lisa [10]

Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.

2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

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3 years ago

What volume will 50.2 grams of co2 (g) occupy at stp?

Genrish500 [490]

Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols

For every 1 mol of gas, there will be
24000 cm^3 of gas

Vol. = 1.14 x 24 dm^3
= 27.36 dm^3

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3 years ago

The melting point of aluminum is about 660 °C. How many degrees above the boiling point of water is this temperature?

Arte-miy333 [17]

560 degrees above water

8 0

2 years ago

What is the theoretical yield of SO3 produced by 8.96 g of S?

Virty [35]

Answer: Theoretical yield of $SO_3$ produced by 8.96 g of S is 33.6 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} S=\frac{8.96g}{32g/mol}=0.28moles$

The balanced chemical equation is:

$2S+3O_2\rightarrow 2SO_3$

According to stoichiometry :

2 moles of $S$ produce = 3 moles of $SO_3$

Thus 0.28 moles of $S$ will produce= $\frac{3}{2}\times 0.28=0.42moles$ of $SO_3$

Mass of $SO_3=moles\times {\text {Molar mass}}=0.42moles\times 80g/mol=33.6g$

Thus theoretical yield of $SO_3$ produced by 8.96 g of S is 33.6 g

7 0

2 years ago

β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa

gtnhenbr [62]

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273.15 +25]=298.15 K$

Putting values in above equation, we get:

$0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L$

The concentration of solute is $2.4278\times 10^{-5} mol/L$

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

$C=\frac{n}{V(L)}$

$n=2.4278\times 10^{-5} mol/L\times 0.137 L$

$n=3.3261\times 10^{-6} mol$

To calculate the molecular mass of solute, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of β‑Galactosidase = $3.3261\times 10^{-6} mol$

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

$3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol$

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

3 0

2 years ago

The airplanes parts and functions​

The airplanes parts and functions