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3 years ago

How many mg of metallic ions are in 1.0 mL of 0.10 M LINO3

Chemistry

1 answer:

Anton [14]3 years ago

8 0

Answer:

0.6941 mg

Explanation:

First we calculate how many LiNO₃ moles there are, using the given concentration and volume:

1.0 mL * 0.10 M = 0.10 mmol LiNO₃

As 1 mol of LiNO₃ contains 1 mol of Li, in the problem solution there are 0.10 mmol of Li (the only metallic ion present).

Now we convert Li milimoles into miligrams, using its atomic mass:

0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg

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A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c

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Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

$E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V$

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq)+2e^-$

Reduction half reaction: $Cu^{+}(aq)+e^-\rightarrow Cu(s)$ ( × 2)

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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

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Putting values in above equation follows:

$E^o_{cell}=0.52-(-0.25)=0.77V$

Hence, the standard potential of the cell is 0.77 V

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3 years ago

How does distillation remove hardness from water

Salsk061 [2.6K]

Answer:

Distillation can be used to soften water as the water evaporates as it is heated and the ions are left behind.

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2 years ago

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Why is the use of high-frequency radio waves a beneficial technological advancement

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Answer:

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3 years ago

What is the pOH of a 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

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3 years ago