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Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)
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The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
Learn more about stoichiometry:
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