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4 years ago

One of the most common elements is Oxygen.

Chemistry

1 answer:

nata0808 [166]4 years ago

4 0

Answer:

Explanation:

- The atomic number of Oxygen is 8.

- The relative atomic mass is approximately 16.

- It has 8 number of protons.

- It has 8 number of neutrons ( Atomic Mass - number of protons).

- Oxygen has 6 valence electrons.

- It has 2 energy shells, two electrons occupy the first energy shell while six electrons occupy the second energy shell.

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What is the empirical formula for C4H6

madreJ [45]

Answer: C2H3

Explanation:simplest whole # ratio

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3 years ago

What will happen if you weighed out a quantity of beans

Ostrovityanka [42]

Answer:

When we weigh beans in this mass ratio, we must obtain the same number of beans.Explanation:

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3 years ago

Suppose a piece of silver jewelry contains 7.49x10^22 atoms of silver (Ag). how many moles of silver are in the jewelry?

zalisa [80]

Answer:

0.124 mol Ag

General Formulas and Concepts:

Atomic Structure

Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 7.49 × 10²² atoms Ag

[Solve] mol Ag

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Step 3: Convert

[DA] Set up: $\displaystyle 7.49 \cdot 10^{22} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})$
[DA] Divide [Cancel out units]: $\displaystyle 0.124377 \ mol \ Ag$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

0.124377 mol Ag ≈ 0.124 mol Ag

Topic: AP Chemistry

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3 years ago

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xz_007 [3.2K]

Alright bet! Say less

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3 years ago

Read 2 more answers

How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?

mr Goodwill [35]

C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2

5 0

4 years ago