They meet all the environmental conditions an organism needs to survive
Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:
3.59x10^21 molecules
Explanation:
1mole of a substance contains 6.02x10^23 molecules.
Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules
1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.
1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.
Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules
Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
Explanation :
Let the volume of sodium benzoate (salt) be, x
So, the volume of benzoic acid (acid) will be, (100 - x)
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:

x = 29.0
The volume of sodium benzoate = x = 29.0 mL
The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL
Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.