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3 years ago

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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

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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Answer:

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The density of a substance can be found by using the formula

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From the question we have

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A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.

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The answer is 3.

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Now, let's implement this in the equation:

<span> $(1/2) ^{n} =0.125$
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A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

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Answer:

The final volume of the sample of gas $V_{2}$ = 0.000151 $m^{3}$

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Put all the values in the above equation we get

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This is the final volume of the sample of gas.

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Answer:

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