Answer:
Explanation:
C = 41.4/12 = 3.43
H = 3.47/1 = 3.47
O = 55.1/16 =3.44
CHO is the skeletal formula (divide each by the lowest number above). The results are close enough to 1 to be 1.
(CHO)_x = 116
C + H + O = 29
(29) _ x = 116
x = 116/29
x = 4
So there area 4 carbons 4 hydrogens and 4 oxygens.
The correct formula is C4H4O4
Answer:
Halophile.
Explanation:
Halophile microorganisms are microorganisms that require very large amounts of NaCl. If the concentration of NaCl is very little, there will be no growth. Ih this case, given that it grows between 5% and 15%, it can be considered a moderate halophile.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
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