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Anuta_ua [19.1K]
2 years ago
15

Help i know whofkfkfkf

Chemistry
1 answer:
Irina18 [472]2 years ago
7 0

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u>

1. A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom. ... Elements that compose the compound are chemically combined. Elements contain only one type of atom.

2. Two ways to add energy to break down a compound are to apply heat and to apply an electric current.

3. The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.

4. In chemistry, the law of definite proportion, sometimes called Proust's law, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

5. compounds and elements.

6.Solid nonmetals

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Which equation shows the correct relationship between mass, force, and acceleration?
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The acceleration on an object is proportional to the resultant force and inversely proportional to the mass of the object  

<h3>Further explanation  </h3>

Newton's 2nd law explains that the acceleration produced by <em>the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  </em>

∑F = m. a  

\large {\boxed {\bold {a = \frac {\sum F} {m}}}

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Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necess
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Answer:

1.242 \times 10^{25}\text{ atoms H}

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}

2. Molecules of B₄H₁₀

\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}

3. Atoms of H

\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}

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