Answer:
Okay so my wifi is slow. How are you!
Explanation:
I don't see an image so I assume it's free points.....thank you! But if there is an image let me know! and Brainliest?? Pretty please!
Answer:
Having as wide a range of organisms as possible.
Hope it helps! :)
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
Answer:
here it is
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solution:
Calculate the molar concentration of the polycyclic aromatic hydrocarbon(PHA)(178.23g/mol),that was found in a well water sample at a concentration of 6.21ppb.Assume the density of the water is 1.00mg/mL