Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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(According to Hess Law)


I don't know what you mean by type of 18, but it is called inorganic chemistry.
<h3>
Answer:</h3>
690 g AgCl
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Writing Compounds
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Limiting Reactant/Excess Reactant
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂
↓
[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂
[Given] 2.4 mol ZnCl₂
[Solve] <em>x</em> g AgCl
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol ZnCl₂ → 2 mol AgCl
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
<u>Step 3: Stoich</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
687.936 g AgCl ≈ 690 g AgCl
The common name to the family member of phylum Echinodermata of marine family is echinoderm. They are usually characterized by a five-fold symmetry, and possess an internal skeleton of calcite plates. They are found at every ocean depth.
The features of all adult echinoderm are:
- They have a five-fold symmetry.
- Body without segmentation.
- Spiny skin.
- Internal skeleton.
- found at every ocean depth.
Answer:
The answer is
<h2>1.0 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 10 g
volume = 10 cm³
It's density is

We have the final answer as
<h3>1.0 g/cm³</h3>
Hope this helps you