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Answer:
Amount of mercury is 1.0*10⁻⁵ g
Explanation:
<u>Given:</u>
Mercury content of stream = 0.68 ppb
volume of water = 15.0 L
Density of water = 0.998 g/L
<u>To determine:</u>
Amount of mercury in 15.0 L of water
<u>Calculation:</u>

where 1 μg (micro gram) = 10⁻⁶ g
0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water
Therefore, the amount of mercury in 15.0 L water would be:

Answer:
The correct answer is 639 L H₂
Explanation:
We use the ideal gas equation:
We have the following data:
n = 28.3 mol
T= 297 K
P= 1.08 atm
R= 0.082 L.atm/K.mol (gas constant)
We introduce the data in the gas equation to calculate the volume (V):
PV=nRT
⇒V =nRT/P = (28.3 mol x 0.082 L.atm/K.mol x 297 K)/(1.08 atm) = 638.2 L ≅ 639 L
Therefore, the correct option is 639 L H₂