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2 years ago

Consider the following reaction, where A and

Chemistry

1 answer:

AlekseyPX2 years ago

4 0

Answer:

Enter your numerical answer to 1 decimal place. Do not include the units of "g". Question: Question 14 1 pts Consider the following reaction, where A and E are hypothetical elements. A+3E --> AE The atomic mass of A is 42.1 amu. The atomic mass of Eis 56.2 amu. If 42.6 g of A fully reacts, how many grams of AE, are expected to form?

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The atomic particles given off as a result of nuclear fusion are?

Vilka [71]

Answer:

I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.

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2 years ago

Read 2 more answers

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,

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

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

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

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

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3 years ago

State the name of the ion which is oxidised in the following half equations. Cathode: Na+ + e– → Na Anode: 2Cl– → Cl2 + 2e–

jenyasd209 [6]

Answer:

hahahahhahhhahahaha

Explanation:

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2 years ago

How many atoms of zirconium are in 0.3521 mol of zirconium?

lora16 [44]

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

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2 years ago

Which element(s) is being reduced in the redox reaction below? 6 nh4clo4 (s) + 10 al(s) 5 al2o3 (g) + 6 hcl (g) + 3 n2 (g) + 9 h

vitfil [10]

The answer is Cl because its oxidation number goes from a +7 down to a -1.

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2 years ago