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Alekssandra [29.7K]

3 years ago

11

what would the total pressure of a mixture of fluorine, chlorine, and bromine gases be if the partial pressure are 2.20 atm, 6.7

0 atm, 0.03 atm respectively

1 answer:

astraxan [27]3 years ago

4 0

Answer:

The total pressure would be 8, 93 atm

Explanation:

We apply Dalton's laws, where for a gaseous mixture, the total pressure (Pt) is the sum of the partial pressures (Px) of the gases that make up the mixture.

Pt= Pxa + Pxb+ Pxc....

Pt=2, 20 atm+ 6, 70 atm+ 0,03 atm= 8, 93 atm

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If an unknown element displays many extremely strong metallic properties, where is it most likely to be located on the periodic

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3 years ago

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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

When a magnesium ribbon is heated in air the product former is heavier.on the other hand when potassium manganate(VII) is heated

vesna_86 [32]

Answer: On heating, Magnesium forms its oxide; while potassium manganate(VII) decomposes

Explanation:

Magnesium Mg, on heating forms Magnesium oxide

2Mg(s) + O2(g) --> 2MgO

Potassium permanganate KMnO4, on heating decomposes to potassium manganate K2MnO4, manganese dioxide MnO2, and Oxygen gas O2.

2KMnO4 --> K2MnO4 + MnO2 + O2

The difference in observation is that, on heating, Magnesium forms its OXIDE as product; while potassium manganate(VII) decomposes, giving OFF most of its constituents and reducing its weight.

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Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

trapecia [35]

Answer:

$CuCl_{2}(aq.)$ will form a precipitate of insoluble $CuCO_{3}$ when aqueous $Na_{2}CO_{3}$ is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and $NH_{4}^{+}$

all salts of sodium are soluble

When $Na_{2}CO_{3}$ is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of $Na_{2}CO_{3}$ into $CuCl_{2}(aq.)$ will result precipitation of insoluble $CuCO_{3}$

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RUDIKE [14]

7 valence electrons in an atom

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