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dsp73
3 years ago
11

A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two

tablets are used to treat 1.00 liter of drinking water, what is the concentration in ppm of iodine in the treated water
Chemistry
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

16\ \text{ppm}

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = 2\times 20=40\ \text{mg}

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = 0.4\times 40=16\ \text{mg}

So, mass of solute is 16\ \text{mg}

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is 1\times 1=1\ \text{kg}=1000\ \text{g}

PPM is given by

\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}

Hence, the concentration of iodine in the treated water 16\ \text{ppm}.

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2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?
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