The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander woul
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<h3>Answer:</h3>
4.42×10^34 molecules/min
<h3>Step-by-step explanation:</h3>Multiply the various factors, along with the unit conversion (60 s/min).
... (3.35×10^25 molecules/L) × (2.2×10^7 L/s) × (60 s/min)
... = (3.35×2.2×60)×10^(25+7) molecules/min
... = 442.2×10^32 molecules/min
... ≈ 4.42×10^34 molecules/min
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The answer to the question is c
