Per hour-68x60=4080
Per day-4080x24=97920
Per week-97920x7=685440
<span>The Keq expression for I2(s) + H2O(l) H + (aq) + I-(aq) + HOI(aq) equilibrium is
H2(g) + I2(g) <----> 2 HI(g).In this Keq expression hydrogen reacts with iodine gives Hydrogen Iodide.HI is a diatomic molecule and it is one of the primary source of iodine which act as a reducing agent.Totally it is the reverse chemical equilibrium reaction.</span>
Answer:
number of moles of the compound 
53 mole
Explanation:
Given that:
The total energy liberated = - 2870 kJ ( here , the negative sign typical implies the release of energy due to the combustion reaction)
The equation of the reaction can be represented as:
The energy needed to synthesize 1 mole of compound X = - 54.1 kJ.mol
Thus;
The total energy = numbers of moles of compound × Energy needed to synthesize 1 mole of compound X
Making the numbers of moles of the compound the subject; we have;
numbers of moles of compound = 
number of moles of the compound = 53.04990 mole
number of moles of the compound 53 mole to two significant figure
Answer:
Decant means "to pour." Kids moving water back and forth between two cups, your dad pouring a bucket of soapy water in the sink, or a wine expert emptying a bottle of wine into a fancy glass container — all of them are decanting liquids.
Explanation:
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.
