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How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106

il63 [147K]

Hey there :)

Question: How many km are in 5.6mm?

=>5.6x10^3

=>5.6x10^-6

=>5.6x10^-3

=> 5.6x10^6

Answer:-

$5.6 \times 10^{ - 6}$

Explanation:-

By using the formula-

$1millimeter = \frac{1}{1000000}$

As 1 with 6 zeros, we convert it into exponential form.

$= > \frac{1}{10^{6} }$

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

$= > 10^{ - 6}$

Now combine with given question.

$= > 5.6 \times 10^{ - 6}$

7 0

2 years ago

How did the uncertainty principle influence thinking about the arrangement of electrons in atoms?

pishuonlain [190]

Answer:

The correct answer is option C

Explanation:

According to Heisenberg's principle "At the instant of time when the position is determined, that is, at the instant when the photon is scattered by the electron, the electron undergoes a discontinuous change in momentum. This change is the greater the smaller the wavelength of the light employed, i.e., the more exact the determination of the position. At the instant at which the position of the electron is known, its momentum therefore can be known only up to magnitudes which correspond to that discontinuous change; thus, the more precisely the position is determined, the less precisely the momentum is known".

Hence, this principle made scientists to realize that electrons could not be located in defined orbits which a contradictory of Bohr's model.

3 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

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

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

4 years ago

when 3.18 g of copper (||) oxide were carefully heated in a stream of dry hydrogen, 2.54 g of copper and 0.72 g of water were fo

gavmur [86]

Answer:

the number of moles of atom is 0.91584

8 0

3 years ago

Imagine you are standing in a kitchen and then

ozzi

You would see mostly compounds because elements make compounds and with these elements making compounds you are most likely not going to see them and since your in the middle of the forest you won't see much mixtures as much as compounds since we create most mixtures. Hope this helps my dude ^^

5 0

4 years ago