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3 years ago

Mg(ClO2)2 and Al(ClO2)3 contain the same number of atoms

1 answer:

7 0

<span>Mg(ClO2)2 has an atom of magnesium, 2 moles of chlorine and 4 moles of oxygen. </span><span>Al(ClO2)3, on the other hand, is composed of one mole of aluminum, one mole of chlorine and six moles of oxygen. The number of molecules of both is equal to 6.022 x10^23 molecules per mole of the compound.</span>

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HELP ME PLEASEE!!!! BRAINLIEST FOR CORRECT ANSWER NO COPY AND PASTE. CLICK ON THE IMAGE

madreJ [45]

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3 0

2 years ago

Read 2 more answers

Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det

Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² * 3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>

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3 years ago

If you wanted to harvest cranberries which separation method would you use

aleksklad [387]

Hello.

If you wanted to harvest cranberries you would need to use flotation.

Have a nice day.

4 0

3 years ago

Describe the process of crystallization.

dalvyx [7]

Crystallization is a process where a solid forms into a substance we call crystal, this happens from precipitating from a solution, freezing, and more

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1 year ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago