Answer:
Ryobi P517 18V Lithium Ion Cordless Brushless 2,900 SPM Reciprocating Saw w/ Anti-Vibration Handle and Tool-Less Blade Changing
Explanation:
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.
First-order reactions are very commonplace. we have already encountered examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.
Rate constant K₁ = 0. 543 /s
T₁ = 25°C
Activation energy Eₐ = 75. 9 k j/mol.
T₂ = 32. 3 °C.
K₂ =?
formula;
log K₂/K₁= Eₐ /2.303 R [1/T₁ - 1/T₂]
putting the value in the equation
K₂ = 0.343 /s
Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s
The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.
Learn more about activation energy here:- brainly.com/question/26724488
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Answer:
Promotes Stellar Formation:
-Increased Gravitational Attraction
-Higher Temperature
Does Not Promote Stellar Formation:
-Decreased Gravitational Attraction
-Lower Temperature
Explanation:
Stars need at least three million kelvins to form, and the gravitational attraction helps form the star in the first place.
