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3 years ago

Due fan 22 at 2.30pm

Chemistry

1 answer:

navik [9.2K]3 years ago

6 0

Answer:

im not sure but i do know you have to look under 5.2

Explanation:

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When a membrane that is permeable to both a solute and water separates water with a high osmolarity from water with a low osmola

slavikrds [6]

<span>The water will diffuse through the membrane until there is an equal concentration of solute and water on both sides.</span>

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3 years ago

Please help!!i NEED to get this right

Lana71 [14]

The correct answer is C

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3 years ago

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How many atoms are in a 1.8 mol sample of Magnesium (Mg)?

Gnom [1K]

Answer:

1.1 × 10²⁴ atoms Mg

General Formulas and Concepts:

<u>Atomic Structure</u>

Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

[Given] 1.8 mol Mg

[Solve] atoms Mg

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.8 \ mol \ Mg(\frac{6.022 \cdot 10^{23} \ atoms \ Mg}{1 \ mol \ Mg})$
[DA] Multiply [Cancel out units]: $\displaystyle 1.08396 \cdot 10^{24} \ atoms \ Mg$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg

5 0

3 years ago

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An 8.42 g piece of metal with a specific heat of 1.020 J g-1 k-1 wasn't heated to an unknown temperature. The metal was then pla

ANTONII [103]

Answer:

$T = 61.344\,^{\textdegree}C$

Explanation:

The heat received by water is equal to the heat rejected by the piece of metal. That is to say:

$-Q_{piece} = Q_{water}$

$(8.42\,g) \cdot \left(1.020\,\frac{J}{g\cdot ^{\textdegree}C} \right) \cdot (T - 20.40\,^{\textdegree}C)= (44.2\,g)\cdot \left(4.187\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (20.40\,^{\textdegree}C-18.50\,^{\textdegree}C)$

$\left(8.588\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20.40\,^{\textdegree}C) = 351.624\,J$

The initial temperature of the piece of metal is:

$T = 61.344\,^{\textdegree}C$

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3 years ago

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In November 1987, a massive iceberg broke loose from the antartic ice mass and floated free in the ocean. The chunk of ice was e

il63 [147K]

<h2>Answer:</h2>

1.58 × 10∧16 pools.

<h3>Explanation:</h3>

Given:

Length of ice berg= 98 miles = 1557716 meters

Width of iceberg = 25 miles = 40233.6 meters

Thickness of iceberg = 750 ft = 230 meters

Volume of water in a swimming pool = 24,000 gallons = 90850 liters

The volume of the ice berg:

Volume = Length . width . thickness

Volume = 1557716 . 40233.6 . 230 = 1,441, 468, 016, 5248 m3 = 1,441, 468, 016, 5248 × 10 ∧3 L.

1 pool contains liters of water: 90850 liters

1,441, 468, 016, 5248 × 10 ∧3 liters contains = 1/90850 . 1,441, 468, 016, 524.8 × 10 ∧3 .

= 1.100 . 1,441, 468, 016, 5248 × 10 ∧3 L.

= 1.58 × 10∧16 pools.

Hence 1.6 × 10∧16 pools will be filled with that chunk of ice.

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3 years ago