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:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
hey there!:
H2S(aq) <=> H⁺(aq) + HS⁻(aq)
K'c = [H⁺][HS⁻]/[H₂S] = 9.5*10⁻⁸
HS⁻(aq) <=> H⁺(aq) + S²⁻(aq)
K"c = [H⁺][S²⁻]/[HS⁻] = 1.0*10⁻¹⁹
H₂S(aq) <=> 2 H⁺(aq) + S²⁻(aq)
Kc = [H⁺]²[S²⁻] / [H₂S]
= [H+][HS⁻] / [H₂S] * [H⁺][S²⁻]/[HS⁻]
= K'c *K"c
= ( 9.5*10⁻⁸ ) * ( 1.0 x 10⁻¹⁹ )
= 9.5*10⁻²⁷
Hope this helps!
Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)
Answer:
709 (With sig figs)
Explanation:
Pressure(1) * volume(1) = Pressure(2) * volume(2)
771.75 mm Hg * unknown = 874.27 mm Hg * 626 mL
771.75 mm Hg * unknown = 547,293.02 mm Hg*mL
Divide both sides by 771.75 mm Hg
Unknown = 709 (With sig figs)....709.158432 (without sig figs)
