Answer: 0.025 moles of nitrogen gas are there in the sample.
Explanation:
According to ideal gas equation:
P = pressure of gas = 1.03 atm
V = Volume of gas = 568 ml = 0.568 L (1L=1000ml)
n = number of moles = ?
R = gas constant =
T =temperature =
0.025 moles of nitrogen gas are there in the sample.
Answer: I think its C or B
Explanation: Hope this was helpful....
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
I think the correct answer would be C. The expression that would best represent a second order rate law would be r =k[X][Y]. Reaction with this rate law are those that depend on the concentration of two first order reactants or a second order reactant.
Answer:
[SO2Cl2] = = 0.015 M
[SO2] = = 0.0027 M
[Cl2] = = 0.0027 M
Q = = = 4.8 × 10−4
No. Q < Kc, so reaction will shift to the right.
Explanation:
