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2 years ago

10

Magnesium hydroxide is only very slightly soluble in water. The reaction by which it goes into solution is: Mg(OH) 2 (s) rightle

ftharpoons Mg^ 2+ (aq)+2OH^ minus (a q) What will happen if O H minus is added to the solution and why?

1 answer:

prohojiy [21]2 years ago

3 0

Answer:

there will a definite decrease in solute solution

Explanation:

acid reaction acting upon negative charge.

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Elemental Li and Na are prepared by electrolysis of a molten salt, whereas K, Rb, and Cs are prepared by chemical reduction.(c)

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Calcium is used to isolate Rb from molten RbX because calcium has a smaller atomic radius than rubidium.

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1 year ago

What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid

vladimir1956 [14]

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Explanation:

On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid $H_2SO_4$ ). The acid medium will produce the <u>hydronium ion</u> ( $H^+$ ). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.

See figure 1

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2 years ago

4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.

ikadub [295]

<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
[DA] Multiply/Divide [Cancel out units: $\displaystyle 0.126923 \ mol \ Au$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.126923 mol Au ≈ 0.127 mol Au

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