E. coli and Salmonella spp.
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
.</span>
Answer:
B. Water will freeze
<em>because</em><em> </em><em>the</em><em> </em><em>latent</em><em> </em><em>heat</em><em> </em><em>of</em><em> </em><em>vapourization</em><em> </em><em>decreases</em><em>.</em>
Answer: C2H2
Explanation: Because each of the lines represent one bond, and because there are three lines (bonds) between the carbons, it means that they are bonded by three bonds, also known as a triple bond.
