The hypothesis is that salt water freezes faster than fresh water.
The dependent variable is time taken for ice to appear.
The independent variable is presence or absence of salt
The constants are the amount of water in each tray, freezing conditions and length of time of exposure to freezing condition.
The control group is the tray to which salt was not added
The experimental group is the tray to which salt was added
The presence of solutes in a solution causes the freezing point depression.
A solution is made up of a solute and a solvent. In the presence of a solute, the freezing point of a pure solvent is decreased. This is because freezing point is a colligative property.
Colligative properties depend on the amount of solute present.
Hence, the pure water freezes faster (ice begin to appear earlier) than the salt water.
The hypothesis put forward in this experiment was found to be invalid by the experiment.
For more about colligative properties, see
brainly.com/question/10323760
When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺
Answer:
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Explanation: I think this is what you are looking for. Hope this helps.
<h3>Answer:</h3>
64 g O₂
<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>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Divide/Multiply [Cancel Units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).
