I am pretty sure it is B. If it turns litmus paper red it is acidic. The other 2 make no sense.
I might be wrong tho.
Hope that helped☺️
mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
a. k=0.256/day
b.sample of Au-198 remains after seven days : 1.67 g
<h3>Further explanation</h3>
1. A half-life of 2.7 days⇒t1/2=2.7 days
The half-life can be expressed in a decay constant( λ)

2. We can use formula : (integrated rate law) :
![\tt ln[A]=-kt+ln[Ao]\\\\ln(\dfrac{Ao}{A})=kt](https://tex.z-dn.net/?f=%5Ctt%20ln%5BA%5D%3D-kt%2Bln%5BAo%5D%5C%5C%5C%5Cln%28%5Cdfrac%7BAo%7D%7BA%7D%29%3Dkt)
Ao=10 g
t=7 days
k=0.256/day

Answer:
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Answer:
es nutricionista el experto en esa area