Answer:
the specimen has 11460 years old
Explanation:
if the live sample has as initial amount of Yo C14, the dead sample will have 0.25Yo of C14.
the rate of decay of radiactive matter over time is
- Y(t) = Yo * e (( - t * Ln2 ) / T)
∴ Y(t) = 0.25Yo; T = 5730
⇒ 0.25Yo = Yo * e (( - t * Ln2 ) / 5730 )
⇒ 0.25 = e (( - t * Ln2 ) / 5730 )
⇒ Ln(0.25) = ( - t * Ln2 ) / 5730
⇒ - 7943.466 = - t * Ln2
⇒ 7943.466 / Ln2 = t
⇒ t = 11460 year
Answer: The ion is 
. The chemical symbol of the noble gas this ion is isoelectronic with Kr.
Explanation:
The given electronic configuration is,
The number of electrons = 2 + 2 + 6 + 2 + 6 = 18 electrons
As the ion has a charge of +2 , that means it has lost two electrons so the the neutral element will have 2 more electrons i.e 18+2 = 20. The element has atomic number of 20 and the element will be Calcium with electronic configuration of
Thus the symbol of ion is . the nearest noble gas with 18 electrons is Krypton with symbol Kr.
Answer:
6.2g of sugar solution contains 72.5g
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
