Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
You’ve not put anything below so I’m not sure what the options are but the relative charge of a proton is +1
I think its an Ionic bond although i am not sure about
A) the independent variable is the variable that doesn’t rely on another variable. For this question the independent would be the days.
b) the dependent variable is the number of bacteria
c) ‘Number of bacteria across a number of days’
Ppm = mass of solute mg / mass of solvent kg
0.008 * 1000 = 8.0 mg ( solute )
1000 / 1000 = 1.0 kg (solvent )
ppm = 8 / 1
= 8.0 ppm
hope this helps!
