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tamaranim1 [39]

3 years ago

5

The atomic radii of a divalent cation and a monovalent anion are 0.62 nm and 0.121 nm, respectively. (a) Calculate the force of

attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).

1 answer:

Zepler [3.9K]3 years ago

3 0

Answer:

$\boxed{8.4\times 10^{-10} \text{ N}}$

Explanation:

The formula for the force exerted between two ions is

$F = \dfrac{kq_{1}q_{2}}{r^{2}} = \dfrac{kz_{1}z_{2}e^{2}}{r^{2}}$

where the Coulomb constant

k = 8.988 × 10⁹ N·m·C⁻²

Data:

z₁ = +2; r₁ = 0.62 nm

z₂ = -1; r₂ = 0.121 nm

Calculations:

r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm

$F =\dfrac{8.988 \times 10^{9} \text{ N $\cdot$ m$^{2} \cdot$ C$^{-2}$} \times 2 \times (-1) \times ( 1.602 \times 10^{-19} \text{ C})^{2}}{(0.741 \times 10^{-9} \text{ m})^{2}}\\\\ = -\mathbf{8.4\times 10^{-10}} \textbf{ N}\\\text{The attractive force between the ions is } \boxed{\mathbf{8.4\times 10^{-10}} \textbf{ N}}$

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93.2 mL of a 2.03 M potassium fluoride (KF) solution is diluted with 3.92 L of water.

pishuonlain [190]

Answer:

0.047 M

Explanation:

Given data:

Volume of KF = 93.2 mL

Molarity of KF = 2.03 M

volume of water added= 3.92 L

Solution:

First of all we will calculate the number of moles of KF.

number of moles = Molarity × volume in litter

number of moles = 2.03 mol/L × 0.0932 L

number of moles = 0.189 mol

Molarity of solution:

Total volume = 3.92 L + 0.0932 L = 4.0132 L

Molarity = number of moles / volume in litter

Molarity = 0.189 mol/ 4.0132 L

Molarity = 0.047 M

5 0

4 years ago

How many moles of water (H2O) are present in a beaker containing 45.9 g H2O? Give your answer to the correct number of significa

ra1l [238]

2.55mol H2O. Good luck! :)

3 0

3 years ago

Read 2 more answers

I really can’t do this don’t understand it

masya89 [10]

Answer:

7.96g, 33.79%

Explanation:

I'll try my best to explain the entire process behind this question ;)

From the question, you can write the reaction

$2H_2O(l)->2H_2(g)+O_2(g)$

Now, there are a few reasons it is like this. Oxygen is a diatomic element, meaning it doesn't and can't exist as just O. It exists as O₂. To balance, this, double the amount of water and hydrogen so there is an equal amount of each element on both sides of the reaction (4 H's, 2 O's on the reactant side, and 4 H's, 2 O's on the product side).

From this we can get a mole-to-mole ratio.

Onto the stoichiometry. Our goal in this is to convert from grams of water to grams of hydrogen, and we do so with a mole to mole ratio.

$71.0gH_2O*\frac{1molH_2O}{18.02g} *\frac{2molH_2}{2molH_2O}* \frac{2.02g}{1molH_2}\\\\ =7.96gH_2$

Basically, what I did was divide by water's molar mass to get moles of water, multiplied by the mole-to-mole ratio (2:2) to get moles of H2, and then multiplied by H2's molar mass to get what should be the amount of H2 produced by the reaction.

For percent yield, you can calculate it is such:

$\frac{Actual}{Theoretical}*100$

Plug the numbers in:

$\frac{2.69g}{7.96g}*100\\\\ =33.79%$

So, the percent yield is 33.79%

3 0

3 years ago

Elements in the same period of the periodic table exhibit similar physical and chemical properties.

Vladimir [108]

Answer:

same valency electrons

Explanation:

example g 1 elements

5 0

3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

4 years ago