i need help ls i thank its book 500 grams
Responda:
+ 0,9kJ / mol
Explicação:
Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;
S (mon.) + O2 (g) -> SO2 (g)
Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1
Dado ∆H2 = -296,8kJ / mol
∆H1 = -297,2kJ / mol
∆H = -296,8 - (- 297,2)
∆H = -296,8 + 297,2
∆H = 297,2-296,8
∆H = + 0,9kJ / mol
Portanto, a mudança na entalpia da equação é + 0,9kJ / mol
Answer:
M
Explanation:
The concentration of the analyte in the 5-mL flask would be M
This is a problem of simple dilution that can be solved using the dilution equation;
C1V1 = C2V2,
where C1 = initial concentration, V1 = initial volume, C2 = final concentration, and V2 = final volume.
<em>In this case, the initial concentration (C1) is not known, the initial volume (V1) is 1.00 mL, the final concentration is 6.97 x 10-5 M, and the final volume is 10.00 mL.</em>
Now, let us make the initial concentration the subject of the formula from the equation above;
C1 = C2V2/V1. Solve for C1 by substituting the other parameters.
C1 = 6.97 x 10-5 x 10/1 = M
Oxygen is a liquid under one atmosphere. the boiling point of a substance is the temperature at which vapor pressure of a liquid is equal to the pressure of the surrounding liquid. the boiling point of oxygen at this case is 90.0k. Therefore the temperature of oxygen is 90-273= -183 celsius scale
Explanation and answer:
The molar volume of a gas is 22.4 L at 0 deg. C.
So the molar mass of the gas is, by proportion,
0.16 g * (22400 mL)/(240 mL) = 14.93 g
The molar mass of (CH4)n = 12+4(1) = 16.
So n = 15.93/16 = 1, or the molecular formula is CH4.
Note: The temperature at which the volume was observed was not given. If 240 cm^3 was observed at 20 deg.C, then the volume at 0 deg.C would be
V=240*(273+0)/(273+20) = 223.6
The molar mass = 0.16*22400/(223.6) = 16.03
which gives n = 16/16.03 = 1 again, but more accurately.
