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enot [183]
3 years ago
10

A laboratory experiment requires 250 millimeters of water for boiling. It also requires 100 millimeters of water for a cooling p

rocess.
Chemistry
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

1.05 L

Explanation:

There is some info missing I think this is the original question.

<em>A laboratory experiment requires 250 millimeters of water boiling. it also requires 100 mills of water for a cooling process. If a student performs the experiment three times, how much total water will the student need? Give your answer in liters.</em>

<em />

Step 1: Calculate the volume required for each experiment.

The volume required is the sum of the volumes used: 250 mL + 100 mL = 350 mL

Step 2: Calculate the volume required for the 3 experiments

We have to multiply the volume required for each experiment by 3.

3 × 350 mL = 1050 mL

Step 3: Convert the volume to liters

We use the relation 1 L = 1000 mL.

1050 mL × (1 L/1000 mL) = 1.05 L

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If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
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Answer:

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General Formulas and Concepts:

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

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

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

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

412.072 g Cl₂ ≈ 412 g Cl₂

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