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3 years ago

5.Name the compound below

Chemistry

2 answers:

Usimov [2.4K]3 years ago

5 0

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neonofarm [45]3 years ago

3 0

Answer:

There isn't anything below, but i can help if you edit the question and put it there

Explanation:

~Bre

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Darya [45]

The answer is B. Continental Drift (Confidently)

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3 years ago

Which of these elements has a smaller atomic #? 1=aluminum 2=beryllium 3=tungsten

enot [183]

2
refer to periodic table for explanation

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3 years ago

What is a polyatomic ion? Consulting p.117 in text, list the formulas and names of at least 10 polyatomic ions. When writing the

aleksley [76]

Answer:

The answer to your question is below

Explanation:

Polyatomic ions are ions composed for more than 1 atom. There are polyanions and polycations.

Polyanions have a negative charge and polycations have a positive charge.

Examples

Polyanions Polycations

acetate CH₃COO⁻ ammonium NH₄⁺¹

bromate BrO₃⁻

chlorate ClO₃⁻

hydroxide OH⁻

nitrate NO₃⁻

nitrite NO₂⁻

sulfate SO₄⁻²

phosphate PO₄⁻³

permanganate MnO₄⁻

We write parentheses before or after a polyatomic ion to emphasize that the oxidation number of the atom which interacts with it affects all the atoms that form part of the polyatomic ion.

4 0

3 years ago

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

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3 years ago

What is the mass percent of a sugar solution when 489 grams of sugar is combined with 877 grams of water?

natka813 [3]

The sugar solution was made by mixing 12 grams of water

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2 years ago