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2 years ago

Please help!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

zimovet [89]2 years ago

3 0

Answer:

Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs.

Explanation:

Brainly pls

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The chemical equation, Cr + Fe(NO3)2 → Fe + Cr(NO3)3, is an example of which type of reaction?

Angelina_Jolie [31]

Answer:

Redox type

Explanation:

The reaction is:

2Cr + 3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃

2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.

If we see oxidation state, we see that chromium changes from 0 to +3

Iron changed the oxidation state from +2 to 0

Remember that elements at ground state has 0, as oxidation state.

Iron is being reduced while chromium is oxidized. Then, the half reactions are:

Fe²⁺ + 2e⁻ ⇄ Fe (Reduction)

Cr ⇄ Cr³⁺ + 3e⁻ (Oxidation)

When an element is being reduced, while another is being oxidized, we are in prescence of a redox reaction.

8 0

3 years ago

What are the molecular formula of 20 elements

baherus [9]

Explanation:

Hydrogen (H)

Helium (He)

Lithium (Li)

Beryllium (Be)

Boron (B)

Carbon (C)

Nitrogen (N)

Oxygen (O)

Fluorine (F)

Neon (Ne)

Sodium (Na)

Magnesium (Mg)

Aluminum (Al)

Silicon (Si)

Phosphorus (P)

Sulfur (S)

Chlorine (Cl)

Argon (Ar)

Potassium (K)

Calcium (Ca)

Hope this is correct and helpful

HAVE A GOOD DAY!

3 0

3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$