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Zarrin [17]
2 years ago
11

The mass of 900. atoms of sodium(Na)

Chemistry
1 answer:
ivanzaharov [21]2 years ago
7 0

Answer:

3.44 × 10⁻²⁰ g Na

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

900 atoms Na

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

<u>Step 3: Convert</u>

<u />900 \ atoms \ Na(\frac{1 \ mol \ Na}{6.022 \cdot 10^{23} \ atoms \ Na} )(\frac{22.99 \ g \ Na}{1 \ mol \ Na} ) = 3.4359 × 10⁻²⁰ g Na

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.4359 × 10⁻²⁰ g Na ≈ 3.44 × 10⁻²⁰ g Na

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Because. just because.
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For each of the following sublevels, give the n and l values and the number of orbitals: (a) 5s; (b) 3p; (c) 4f
OverLord2011 [107]

Answer:

(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1

(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3

(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Sublevel number, 0 ≤ l ≤ n − 1

So,

(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1

(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3

(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals =  2l +1 = 7

6 0
3 years ago
. an ionic compound is composed of 34.95 g of iron and 15.05 g of oxygen. find the empirical formula of this compound.
Contact [7]

Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.

Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.

mass of iron in compound = 34.95 g

molar mass of iron = 55.8 g

mass of oxygen in compound = 15.05 g

molar mass of oxygen = 32 g

number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron

number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1

number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5

the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1

Hence , the required empirical formula of iron compound is FeO.

To learn more about Emiprical formula, refer:

brainly.com/question/1439914

#SPJ4

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Answer:

= 15.51 mL

Explanation:

Here's is the reaction:

2HgO(s) ⇒ 2 Hg(s)+O₂(g)

In this reaction 2mol HgO =  1mol O₂

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so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂

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