How do I do this Phenomenal Based writing: Dont mess up the cookies worksheet? I need the answers asap please!!

xxTIMURxx [149]

Answer:

It is in violation of the Heisenberg Uncertainty Principle. The Bohr Model considers electrons to have both a known radius and orbit, which is impossible according to Heisenberg. ... The Bohr Model does not account for the fact that accelerating electrons do not emit electromagnetic radiation

6 0

2 years ago

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consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.

iogann1982 [59]

Answer:

$pH=10.5$

Explanation:

Hello,

In this case, the dissociation of the given weak acid is:

$HA\rightleftharpoons H^++A^-$

Therefore, the law of mass action for it turns out:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction extent is:

$1x10^{-23}=\frac{x*x}{0.1-x}$

Thus, by solving with the quadratic equation or solver, we obtain:

$x=31.6x10^{-12}M$

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

$pH=-log(-31.6x10^{-12})\\pH=10.5$

Regards.

5 0

2 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

Titration Volume & Concentration

polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

$0,0171L*\frac{0,161moles}{L}$ = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

$2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}}$ = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

$5,5062x10^{-3}molHCl*\frac{1L}{0,304mol}$ = 0,0181L≡18,1mL

I hope it helps!

4 0

2 years ago

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0

2 years ago

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