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3 years ago

Oxygen is an example of a(n)

Chemistry

1 answer:

stepladder [879]3 years ago

7 0

Answer:

C. element. oxygen is an element

Explanation:

oxygen is element #8

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Can anyone do my Ap environmental science package I will pay you

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Answer:

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2 years ago

What is the study of the spectra produced from the emission and absorption of electromagnetic radiation? study of absorption spe

skelet666 [1.2K]

Answer the correct answer out of the four is option c spectroscopy

Explanation- The interaction between electromagnetic radiation and matter studied. This study is commonly known as spectroscopy. It can also be named as study of absorption spectra or emission spectra.

The former spectrum is formed when energy is absorbed from Photons or light energy by electrons while the latter spectrum is formed due to a wavelength of light that is released when electrons jump from higher to lower level.

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3 years ago

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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

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sweet [91]

Answer:

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Explanation:

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3 years ago

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Solve if an element has a charge of 2+, how many more protons does it have than electrons? protons

Effectus [21]

To determine how many protons are present, simply know or use the following equation

X = number of protons
Y = number of electrons
Z = formal charge of the element

X - Y = Z = 0 for a neutral atom

X - Y = 2

Then to solve for x

X = Y + 2.

This means that we would need an additional 2 more protons than electrons.

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2 years ago