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3 years ago

Which one is the odd one out between melting and evaporating

Chemistry

1 answer:

Kaylis [27]3 years ago

8 0

Evaporating is the right one

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The difference between paper chromatography and thin layer chromatography is that

Lera25 [3.4K]

The basic difference between thin layer chromatography (TLC) and paper chromatography (PC) is that, while the stationary phase in PC is paper, the stationary phase in TLC is a thin layer of an inert substance supported on a flat, unreactive surface. ... Paper chromatography is performed using paper.

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2 years ago

At equilibrium, Keq for the following reaction is equal to 3.5: SO2 + NO20 È SO3 + NO a) Is a beaker containing 2.5 M SO2, 2.0 M

liraira [26]

Keq=[SO3][SO2] /[SO2][NO2]

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3 years ago

The ground state of an electron is the least stable energy state of an atom

kari74 [83]

Answer:

electron configuration

Explanation:

The arrangement of electrons in the atomic orbitals of an atom is called the electron configuration. Electron configurations can be determined using a periodic table.

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3 years ago

All of the following are indicators of chemical change except…

Oduvanchick [21]

D. All of the above are indicators of chemical change

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3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

2 years ago