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Sever21 [200]
3 years ago
6

The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because the

y can travel farther. Which radioisotope is the best to use?

Chemistry
2 answers:
adoni [48]3 years ago
8 0

Answer:

D. plutonium-239

Explanation:

BaLLatris [955]3 years ago
4 0

Answer : The radioisotope which has longest half-life and can be best to be used in powering planet and space exploration vehicle and travel farther will be Plutonium - 239 as it has the half life of 24065 years. Which in comparison to other radioisotopes is longest.


Hence, Plutonium - 239 can be the best suitable radioisotope for powering the planet and space exploration vehicle to travel farther.

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Which describes interactions between substances and stomata during photosynthesis?
Mariulka [41]
The interaction between the substances and stomata is a chemical reaction, light energy strikes at the chlorophyll molecules, whose electrons gets excited to a state of higher energy, and they come back to their state of lower energy by emission of energy, which is accepted by a chain of acceptors, and energy is generated. 
8 0
3 years ago
Someone help answer this for me tysm
Ghella [55]
Energy, kinetic, potential
8 0
3 years ago
Read 2 more answers
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hammer [34]

Answer:

B is the correct answer

Explanation:

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7 0
3 years ago
a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses
Sergio039 [100]

Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

  • Initial conditions:

P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa

3 0
2 years ago
A 3.31-g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.53 L. T
Dahasolnce [82]

Answer:

0.486atm is the pressure of the cylinder

Explanation:

1 mole of Pb(NO₃)₂ descomposes in 4 moles of NO2 and 1 mole of O2. That is 1 mole descomposes in 5 moles of gas.

To find the pressure of the cylinder, we need to find moles of gas produced, and using general gas law we can determine the pressure of the gas:

<em>Moles Pb(NO₃)₂ and moles of gas:</em>

3.31g * (1mol / 331g) = 0.01 moles of Pb(NO₃)₂.

That means moles of gas produced is 0.05 moles.

<em>Pressure of the gas:</em>

Using PV = nRT

P = nRT/V

<em>Where P is pressure (Incognite)</em>

<em>V is volume (2.53L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature (300K)</em>

And n are moles of gas (0.05 moles)

P = 0.05mol*0.082atmL/molK*300K / 2.53L

P = 0.486atm is the pressure of the cylinder

3 0
3 years ago
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