The mass of pentane the student should weigh out is
The density of pentane is 0.626 gcm-3
To calculate the mass of pentane following expression is used,
(Density is defined as the mass divide by volume)
Density = mass / volume
mass of pentane = Density of pentane * Volume of pentane
mass of pentane = 0.626 gcm-3 * 45.0 mL
= 28.17 g
Here the unit of mass of pentane is g,
However the unit of density is gcm-3 and unit of volume is mL i.e. cm3
Hence, Mass = gcm-3 * cm3
Mass = g
The mass of pentane the student should weigh out is 28.17g
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Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
The answer is D, because "Animals without a backbone are invertebrates."
