Answer:
The solute is the substance being dissolved.
The solvent is the substance dissolving the solute.
Therefore, the salt is the solute and the water is the solvent.
Explanation:
The salt is the solute.
he required empirical formula based on the data provided is Na2CO3.H2O.
<h3>What is empirical formula?</h3>
The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.
We have the following;
Mass of sodium = 37.07-g
Mass of carbonate = 48.39 g
Mass of water = 14.54-g
Number of moles of sodium = 37.07-g/23 g/mol = 2 moles
Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole
Number of moles of water = 14.54/18 g/mol = 1 mole
The mole ratio is 2 : 1: 1
Hence, the required empirical formula is Na2CO3.H2O
Learn more about empirical formula : brainly.com/question/11588623
<h3>
Answer:</h3>
25.4 g CH₄
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.58 mol CH₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
25.359 g CH₄ ≈ 25.4 g CH₄
Answer:
Metals consist of giant structures of atoms arranged in a regular pattern. The electrons from the outer shells of the metal atoms are delocalised , and are free to move through the whole structure. This sharing of delocalised electrons results in strong metallic bonding .
Explanation:
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pH solution = 8.89
<h3>Further explanation</h3>
Given
The concentration of HBr solution = 1.3 x 10⁻⁹ M
Required
the pH
Solution
HBr = strong acid
General formula for strong acid :
[H⁺]= a . M
a = amount of H⁺
M = molarity of solution
HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1
Input the value :
[H⁺] = 1 x 1.3 x 10⁻⁹
[H⁺] = 1.3 x 10⁻⁹
pH = - log [H⁺]
pH = 9 - log 1.3
pH = 8.89