Answer:
Obtención. El carbono se encuentra - frecuentemente muy puro - en la naturaleza, en estado elemental, en las formas alotrópicas diamante y grafito. El material natural más rico en carbono es el carbón (del cual existen algunas variedades). Grafito: Se encuentra en algunos yacimientos naturales muy puro.
We know that:
mass = density x volume
The volume of the total mixture is 0.1 m³
Let the first liquid be A and the second be B
Mass Total = Mass A + Mass B
800 x 0.1 = 1500Va + 500(0.1 - Va)
30 = 1000Va
Va = 0.03 m³
Vb = 0.1 - 0.03 = 0.07 m³
Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
- See charge on ion is -1 .
Hence it has taken 1 electron
Now first look at EC of Fluorine(F)
- Now one electron added .hence no of electrons is 10now
Look at the EC
Or
![\\ \bull\sf\dashrightarrow [He]](https://tex.z-dn.net/?f=%5C%5C%20%5Cbull%5Csf%5Cdashrightarrow%20%5BHe%5D)
Option C is correct.
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
