Question

какой объем кислорода при (н.у) выделится при разложении 0.5 моль оксида ртути

Answer 1

2HgO=2Hg+O2. V(HgO за уравнениям)=44,8л/моль; V(O2)=22,4л/моль; из этого V(O2)=(0,5*22,4)/44,8=0,25л.

Answer 2

I can help you with your question.

The decomposition of mercury oxide is a chemical reaction that produces oxygen and mercury. The balanced equation for this reaction is:

$$2HgO \rightarrow 2Hg + O_2$$

To find the volume of oxygen that is produced at normal conditions, we need to use the ideal gas law:

$$PV = nRT$$

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At normal conditions, the pressure is 1 atm, the temperature is 273 K, and the gas constant is 0.0821 L atm / K mol. We can plug in these values and solve for V:

$$V = \frac{nRT}{P}$$

$$V = \frac{0.25 \times 0.0821 \times 273}{1}$$

$$V = 5.6 L$$

Therefore, the volume of oxygen that is produced at normal conditions is 5.6 liters.Ответь+подробно.+какой+объем+кислорода+при+(н.у)+выделится+при+разложении+0.5+моль+оксида+ртути