Meat contamination because humans do not eat air or fertilizer so meat contamination is the only logical answer.
Explanation:
Pineapple juice often shows an unstable cloud and produces a solid precipitate that is not very attractive for consumers. Cloud stabilization by pectin addition is permitted by EU and Codex standards to counteract this effect. This additive must be labeled and its content should not exceed fixed maximum standards (Website of AIJN Code of Practice). Determination of water-soluble pectins by IFU method 26 (Website of International Fruit and Vegetable Juice Association) can be used for control of this parameter. Pectin addition to pineapple juice or juice concentrate, etc. may also be detected after its isolation by 13C isotopic analysis (Hammond, 2006) as explained later.
Carbon
calcuim
iron
hydrogen
iodine
nitrogen
oxygen
phosphorus
sulfur
The letters are the symbols for those elements
When the balanced reaction equation for the reaction of methane gas and O2 is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
we will use the ideal gas equation to get the moles of each gas:
1) moles of methane:
PV= n RT
when p is the pressure =1.65 atm
V is the volume = 2.8 L
R is the ideal gas constant = 0.0821
T is the temperature in Kelvin = 25+ 273= 298 K
∴n of CH4 = 1.65 * 2.8 / 0.0821 * 298 K = 0.189 mole
2) moles of O2:
PV = nRT
when P = 1.25 atm
and V= 35L
and R = 0.0821 L atm / mol K
T = 31 + 273 = 304 K
so, by substitution:
n of O2= 1.25 atm * 35 L / 0.0821 * 304 = 1.75 mol
from the balanced reaction equation, we can see that the molar ratio between O2 and CH4 is 2: 1
∴ moles of O2 = 2 * .0189 moles of methane = 0.378 mole
so the O2 exists in excess and CH4 is the limiting reactant
when 1 mole of CH4 → 1 mole ofCO2
1 mol of CO2 → 22.4 L CO2
volume of CO2 at STP =( 0.189 molesCH4 /1)*(1mol CO2)/(1 mole CH4)*(22.4L CO2)/(1molCO2) = 4.23 L
then we will use this formula to get the correct value of V of CO2 at laboratory conditions:
P1V1 /T1 = P2V2/T2
when at STP conditions:
P1= 1 atm
V1 = 4.23 L
T1 = 273
and the laboratories conditions are:
P2 = 2.5 atm
T2 = 125 + 273 = 398 K
∴ V2 = 1 atm * 4.23 L *398 / 273 * 2.5 atm
∴ V2 = 15.4 L
∴ the volume of CO2 formed at pressure of 2.5 atm and T = 125°C is 15.4 L
