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torisob [31]
2 years ago
7

What happens if the solvent solute attraction is greater than the salute solute attraction when to substances are mixed

Chemistry
1 answer:
dolphi86 [110]2 years ago
5 0

Answer:

What happens if the solvent-solute attraction is greater than the solute-solute attraction when two substances are mixed? The solute particles are pulled apart and dispersed throughout the particles of the solvent, which holds the particles of the solute in solution.

Explanation:

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200 g of solid sodium hydroxide (NaOH) are dissolved in 750 mL of water. What is the concentration of this solution?
Anastaziya [24]

Answer:

6.67 M

Explanation:

Molarity = \frac{mol}{L}

<em>Convert 200g NaOH to moles. Convert 750 mL to L.</em>

200 g NaOH x (1 mol/39.998 g) = 5.00025... mol NaOH

750 mL x (1 L/1000 mL) = 0.750 L

<em>Substitute values into the equation.</em>

Molarity = \frac{5.00025}{0.750}

Molarity = 6.667... M

Molarity = 6.67 M

8 0
2 years ago
How many kilojoules are released when 2.25 mol of H2O2 reacts?
Stels [109]

Answer:

230.3 kJ

Some chemical reactions show release of heat while some show absorption of heat. On these bases, the reaction is classified as either endothermic or exothermic. Negative sign of ΔH means exothermic reaction.

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3 years ago
Where two air masses and do not mix
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When two air masses meet together, the boundary between the two is called a weather front. At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.
3 0
3 years ago
Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
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3 years ago
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