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mixer [17]
3 years ago
15

A geochemist in the field takes a 25.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X .

He notes the temperature of the pool, 15.° C , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 450. mL . Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.300 g . Using only the information above, can you calculate the solubility of X in water at 15.° C ? yesno If you said yes, calculate it. Be sure your answer has a unit symbol and 3 significant digits. ×10μ
Chemistry
1 answer:
san4es73 [151]3 years ago
4 0

Answer:

1- Yes, we can calculate the solubility of mineral compound X.

2- 0.012 g/mL.

Explanation:

<em>1- Using only the information above, can you calculate the solubility of X in water at 15.0 °C? </em>

The information available is:

The volume of water sample = 25.0 mL.

Weight of the mineral compound X after evaporation, drying, and washing = 0.30 g.

∴ Yes, we can calculate the solubility of mineral compound X.

<u><em>2-  If you said yes, calculate it.</em></u>

∵ 25.0 mL of water sample contains → 0.30 g of the mineral compound X.

∴ 1.0 mL  of water sample contains → ??? g of the mineral compound X.

1.0 ml of water sample will contain (0.3 g/25.0 mL) 0.012 g.

<em>∴ The solubility of the mineral compound X in the water sample is</em> <u><em>0.012 g/mL.</em></u>

<u><em></em></u>

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What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
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<h2><u>Explanation:</u></h2>

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2.25g x 1 mole / 28.0g = 0.08036 moles = m

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<u>Step 1: Define</u>

[Given] 1.51 × 10²⁶ atoms Xe

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

Avogadro's Number

<u>Step 3: Convert</u>

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<u>Step 4: Check</u>

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250.747 mol Xe ≈ 251 mol Xe

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