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3 years ago

An aerospace engineer is designing a rocket to have three stages. Which is

Chemistry

2 answers:

natima [27]3 years ago

5 0

Answer:

A. To simplify the design of the rocket.

Explanation:

Correct answer for APEX quiz!

valina [46]3 years ago

4 0

Answer:

A not reasonable explanation for an aerospace engineer designs a rocket to have three stages is:

A. To simplify the design of the rocket

Explanation:

A three stages rocket permits to separate the fuel compartments. The rocket must be launched with a huge amount of fuel to last the entire travel. Then the initial mass of the rocket and the fuel is considerably large.

As the rocket flies, part of the fuel is consumed and so some compartements may be separated and removed decreasing the mass of the rocket as it flies (option D. is a reasonable explanation), so the rocket only carries the necessary mass (statement B. is a reasonable explanation).

As the rocket moves away from Earth, the gravitational force decreases and, in the space, the amount of thrust needed is lesser, which the aerospace engineers take into account for the design (choice C. is a reasonable explanation).

The only not reasonable explanation to design a three stage rocket is to simplify the design of the rocket: designing a rocket with three modules each with different features, functions, and the detachment mechanisms brings lot of complex problems to solve. So, option A. is the correct answer.

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What is the atomic mass of one mole of H? g/mole

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Read 2 more answers

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

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

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$