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3 years ago

List the three great pioneers of rocket science and their contributions.

2 answers:

6 0

Here are the three great pioneers of rocket science and some of their contributions:
1. Konstantin Tsiolkovsky - a Russian teacher who realized that liquid fuel instead of solid fuel is needed for rockets to fly outside of Earth's gravity
2. Robert Goddard - an American scientist who created a new system for rockets that cooled that gases that were leaving them which improved the way rockets work significantly
3. Hermann Oberth - a German scientist who created the rocket V-2 which was used in the WWII as a missile bomb

Yanka [14]3 years ago

3 0

Answer:

Here are the three great pioneers of rocket science and some of their contributions:

1. Konstantin Tsiolkovsky - a Russian teacher who realized that liquid fuel instead of solid fuel is needed for rockets to fly outside of Earth's gravity

2. Robert Goddard - an American scientist who created a new system for rockets that cooled that gases that were leaving them which improved the way rockets work significantly

3. Hermann Oberth - a German scientist who created the rocket V-2 which was used in the WWII as a missile bomb

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Explanation:

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A sample of gas occupying 350 mL at 25 C is cooled to -25 C. What volume will it occupy if the pressure is held constant?

leonid [27]

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3 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

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3 years ago

What shape does N2H4 have? (linear, bent, planar, pyramid, tetrahedral, or none)

Salsk061 [2.6K]

The N2H4 bond angle will be probably 109 degrees. Since, well, it has a bent trigonal pyramidal geometry.

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3 years ago

In the gaseous state, chlorine exists as a diatomic molecule Cl2 (Molar mass = 70.9 g/mol). Calculate the number of moles of chl

Annette [7]

Answer:

3.67 mol Cl

Explanation:

We need to convert g of Cl 2 to moles of Cl. First we divide 130 gCl2 by the molar mass (70.90 gCl2/mol) to find out how many moles of Cl2 do we have.

130 gCl2 x $\frac{1 mol Cl2 }{70.90 gCl2}$ = 1.83 mol Cl2

Then we need to convert 1.83 mol de Cl2 to moles of Cl. We have 2 moles of Cl in every Cl2 molecule so we just need to multiply by 2.

1.83 molCl2 x $\frac{2 molCl}{1 molCl2}$ = 3.67 molCl

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3 years ago

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Alla [95]

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