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anygoal [31]
2 years ago
8

In a hydrogen bomb, hydrogen is converted into:

Chemistry
1 answer:
Alona [7]2 years ago
8 0

Answer:

helium ................

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A hot air ballon contains 270L of helium at 24C and 845 mmHg. WHat will be the volume of the ballon at -50c and pressure of 0.73
Julli [10]

Using the ideal gas law,

<em>PV</em> = <em>nRT</em>

where <em>R</em> = 0.08206 L•atm/(mol•°K), solving for <em>n</em> gives

<em>n</em> = <em>PV</em>/(<em>RT</em>)

<em>n</em> = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))

Convert the given temperature to °K and the given pressure to atm:

24 °C = (273.15 + 24) °K ≈ 297.2 °K

(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm

Then the balloon contains

<em>n</em> = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))

<em>n</em> ≈ 12.3 mol

of He.

Solve the same equation for <em>V</em> :

<em>V</em> = <em>nRT</em>/<em>P</em>

<em />

Convert the target temperature to °K:

-50 °C = (273.15 - 50) °K = 223.15 °K

Then the volume under the new set of conditions is

<em>V</em> = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)

<em>V</em> ≈ 306 L

7 0
2 years ago
Condensation point AND freezing point of oxygen in C and F?
aliina [53]

Answer:54.36 K (−218.79 °C; −361.82 °F)

Explanation:

7 0
3 years ago
The bullet train was traveling at a speed of 200mi/hr for 2 hours. What
viva [34]

Answer:

Distance is speed x time

So 200 x 2 =400mi

7 0
3 years ago
Which energy changes in the enthalpy of solution are endothermic, and which are exothermic?
snow_tiger [21]
When energy is needed in order for the reaction to happen, then that reaction is known as endotermic. When the reaction has as a result energy then it is exotermic. An example of an endotermic reaction would be photosynthesis, for an exotermic: combustion.
8 0
3 years ago
The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equatio
Bogdan [553]

Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

                C₁₂H₂₂O₁₁  +  12 O₂  ⇒   12 CO₂  +  11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

6 0
3 years ago
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