Answer: assume pathogens are present and treat the samples accordingly
Explanation:
When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.
This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.
Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
