Answer:
12.5%
Explanation:
Initial percentage of carbon 14 = 100%
Final percentage = ?
Time passed = 17.1 * 1000 years = 17100 years
Half life of carbon 14 = 5,730 years.
So how many Half lives are in 17100 years?
Number of Half lives = Time passed / Half life = 3.18 ≈ 3
First Half life;
100% --> 50%
Second Half life;
50% --> 25%
Third Half life ;
25% --> 12.5%
Answer:
CH4 +2 O2 — CO2 +2 H2O
Now we see that for 1 mol i.e. 16 grams of methane results in 1 mol of CO2 or 44 grams of CO2.
That means for 3 moles of methane , we will obtain 3 moles of CO2 OR for 48 (3*16) grams of CO2 , we will obtain (44*3) 132 grams of CO2 .That's it….
Explanation:
hey .dude hope the answer was helpful ....
In order to find your answer you need to be <span>measuring entropy, so you will be using the following formula:
</span><span>delta S= S of (N2H4) + S of ( H2) - [2( S of NH3)]
</span>Hope this is very useful for you
Answer:
Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²
Explanation:
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²
Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48
Ca. #Substrate e⁻ = 6F = 6(8) = 48
#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs
#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs
BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²
XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)
Water is an amphoteric compound. This means it could be a base or an acid, depending on the substance it is to be reacted with. In this case, water is a base because HF is an acid. Now, if the reactant is an acid, its form after the reaction is called the conjugate pair. Since HF became F⁻, <em>the acid-conjugate base pair is: HF and F⁻.</em>
