<span>Heavy metals like mercury enter waterways by industrial dumping and poor regulatioin of effluent, and they also enter soil through a similar manner, in which waste is disposed of imporperly. Another source of heavy metals are the gases leaving industry carrying these metals. The metals fall as a solid on to soil and water ways. Therefore, the answer is D.</span>
<span>Pre-1982 definition of STP: 37 g/mol
Post-1982 definition of STP: 38 g/mol
This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using.
First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume.
Pre-1982: 2.1 / 22.414 = 0.093691443 moles
Post-1982: 2.1 / 22.71098 = 0.092466287 moles
Now determine the molar mass. Simply divide the mass by the moles. So
Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol
Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol
Finally, round to 2 significant figures. So
Pre-1982: 37 g/mol
Post-1982: 38 g/mol</span>
Answer:
<u><em>I HOPE IT WILL BE YOUR ANSWER:</em></u>
Explanation:
As given
60 min = 50 gm (1)
then we know half-life mean half amount decay time
so we can write as the half of 200 is 100 gm hence
T 1/2 = 100 (2)
solving these two equation by cross multiplication we will get
T 1/2 = 120 min
<em><u>THANKS FOR ASKING QUESTION</u></em>
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
Hey there!
Mass = 5.6 Kg
Volume =8.2 L
D = m / V
D = 5.6 / 8.2
D = 0.6829 Kg/L
hope this helps!
