in one of the routine analyses, dr entropy found that the concentration of phosphate had increased from 0.0090 mg/l to 0.220 mg/l. this will require a lengthy investigation and detailed report into the causes. being the master of disorder, he considers an act of serious scientific misconduct - diluting the sample to reduce phosphate concentration to the maximum allowable level of 0.100 mg/l, and hoping that no-one notices.

if the volume of the original sample with concentration 0.220 mg/l is 10.0 ml, what should be the final volume of solution in order to dilute it to 0.100 mg/l?

The final volume should be 22 mL

Explanation:

For this problem, we will use the dilution equation:

C1*V1 = C2*V2

Step 1: Data given

with C1 = the initial concentration C1 = 0.220 mg/L

with V1 = the initial volume = 10 mL = 10 * 10^-3 L

with C2 = the final concentration = 0.100 mg/L

with V2 = the final volume = TO BE DETERMINED

Step 2: Calculating the final volume

C1*V1 = C2*V2

0.220 mg/L * 10*10^-3 L = 0.100 mg/L * V2

V2 = (0.220 mg/L * 10*10^-3 L) / 0.100 mg/L

V2 =0.022 L = 22 mL

The final volume should be 22 mL

the answer would be b i believe

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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