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stellarik [79]
3 years ago
8

Cs2O + H2O combination

Chemistry
1 answer:
patriot [66]3 years ago
8 0
Cs2O + H2O = CsOH

Hope this helps!
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If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w
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Answer:

The pressure of the gas increased (if temperature remained constant).

The Boyle's law supports this observation.

Explanation:

The initial measurements of the gas are given as;

volume = 100 L

Pressure = 300 kpa

The second measurement is given as;

Volume = 75 L

The second reading implies that the volume of the gas has decreased. If the temperature of the gas remained constant, then the pressure must have increased according to the Boyle's law;

At constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.

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What gives the gem amethyst its purplish color?
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Fatima is designing an advanced graphics card for a next-generation video game system. She needs to be able to
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If you throw a ball into the air, the force of ________ makes the ball fall back toward Earth.
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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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