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3 years ago

The radius of an atom of gold (Au) is about 1.35 Å.

Chemistry

2 answers:

Degger [83]3 years ago

8 0

Answer : The number of gold atoms will be, $3.52\times 10^7$

Explanation :

First we have to determine the diameter of an atom of gold.

$Diameter=2\times Radius$

Given :

Radius of an atom of gold = $1.35\AA$

$Diameter=2\times 1.35\AA=2.7\AA$

Conversion used : $(1\AA=10^{-7}mm)$

$Diameter=2.7\AA=2.7\times 10^{-7}mm$

Now we have to calculate the number of gold atoms.

$\text{Number of gold atoms}=\frac{\text{Span length}}{\text{Diameter of an atom of gold}}$

$\text{Number of gold atoms}=\frac{9.5mm}{2.7\times 10^{-7}mm}=3.52\times 10^7$

Therefore, the number of gold atoms will be $3.52\times 10^7$

Eduardwww [97]3 years ago

6 0

Two radius of an atom is equal to the diameter. Adding up all the diameter of the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms

Hope this answers the question. Have a nice day.

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Enter an equation to show how h2po3− can act as a base with hs− acting as an acid. Express your answer as a chemical equation. I

vladimir2022 [97]

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To show the acid - base relationship between H2PO3^- and HS^-, we have the equation;

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

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After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

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Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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