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Oliga [24]
3 years ago
15

Calculate the volume of carbon dioxide and water vapour produced and the volume of oxygen remaining, when 20.0 dm3 of propane re

acts with 120.0 dm3 of oxygen. All gas volumes are measured at the same temperature and pressure.

Chemistry
1 answer:
laiz [17]3 years ago
6 0

Answer:

80cm3 of water, and 60cm3 carbon IV oxide is formed while 20cm3 of oxygen is left unreacted.

Explanation:

From Gay-Lussac's law, there are five volumes of oxygen, 1 volume if propane, 4 volumes of water and three volumes of CO2. Applying this shows the reacting volumes as we have in the image attached, hence the volumes left after reaction.

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What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The
Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}

We are given:

Mass of _{19}^{40}\textrm{K} = 39.963998 u

Mass of _{20}^{40}\textrm{Ca} = 39.962591 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(39.963998-39.962591)=0.001407u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.001407u)\times c^2

E=(0.001407u)\times (931.5MeV)    (Conversion factor:  1u=931.5MeV/c^2  )

E=1.3106MeV

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

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