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kolbaska11 [484]

3 years ago

14

The factor that affects how easy and electron can be removed from an Atom is the

1 answer:

zheka24 [161]3 years ago

3 0

The ionization energy

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a

trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 + 2 HCl --> MnCl2 + H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)

= 4.86 x 10^-3 mole

this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L

thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M

Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4

6 0

3 years ago