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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`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i

Andrew [12]

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

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T₂ = ?

Using above equation as:

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$T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K$

The conversion of T(K) to T( °C) is shown below:

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3 years ago

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

Answer: The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

Explanation:

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$