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How do nuclear fusion and combustion relate?

RUDIKE [14]

Answer:The key to remember is that carbon combustion requires oxygen. As soon as there is no oxygen left, carbon combustion stops. In nuclear fusion, the nuclei of atoms are fused together to make new, bigger nuclei.

Explanation:

6 0

3 years ago

A gas mixture consists of 4 kg of O2, 5 kg of N2, and 7 kg of CO2. Determine (a) the mass fraction of each component, (b) the mo

kondor19780726 [428]

Answer:

See explanation.

Explanation:

Hello

(a) In this case, one uses the following formulas, which allow to compute the mass fraction of each component: $\% O_2=\frac{m_{O_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{4kg}{4kg+5kg+7kg}*100\%=25\%O_2\\\% N_2=\frac{m_{N_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{5kg}{4kg+5kg+7kg}*100\%=31.25\%N_2\\\% CO_2=\frac{m_{CO_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{7kg}{4kg+5kg+7kg}*100\%=43.75\%CO_2$

(b) For the mole fractions, it is necessary to find all the components' moles by using their molar mass as shown below:

$n_{O_2}=4kgO_2*\frac{1kmolO_2}{32kgO_2} =0.125kmolO_2\\n_{N_2}=5kgN_2*\frac{1kmolN_2}{28kgN_2} =0.179kmolN_2\\n_{CO_2}=7kgCO_2*\frac{1kmolCO_2}{44kgCO_2} =0.159kmolCO_2$

Now, the mole fractions:

$x_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.125kmolO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=27\%O_2\\x_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.179kmolN_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=38.7\%N_2\\ x_{CO_2}=\frac{n_{CO_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.159kmolCO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=34.3\%CO_2$

(c) Finally the average molar mass is computed considering the molar fractions and each component's molar mass:

$M_{average}=0.27*32g/mol+0.387*28g/mol+0.343*44g/mol=34.57g/mol$

And the gas constant:

$Rg=0.082\frac{atm*L}{mol*K}*\frac{1mol}{34.57g}=0.00237\frac{atm*L}{g*K}$

Best regards.

8 0

4 years ago

Help it’s for science

Karolina [17]

Answer:

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Explanation:

6 0

3 years ago

How many grams of water can be vaporized by 2.153 kJ of energy?

Alexxandr [17]

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