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2 years ago

What is the molarity of a solution if 5 moles of solute are dissolved in 8 L of solvent.

Chemistry

1 answer:

tankabanditka [31]2 years ago

6 0

Answer:

0.625mol/L

Explanation:

Given parameters:

Number of moles of solute = 5moles

Volume of solvent = 8L

Unknown:

Molarity = ?

Solution:

Molarity is the number of moles of solute per unit volume of solution. It is mathematically expressed as

Molarity = $\frac{number of moles }{Volume}$

Molarity = $\frac{5}{8}$ = 0.625mol/L

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How many atoms of Iodine are in a 12.75g sample of CaI2?

Inessa [10]

Answer is: 5.22·10²² atoms of Iodine.

m(CaI₂) = 12.75 g; mass of calcium iodide.

M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.

n(CaI₂) = m(CaI₂) ÷ M(CaI₂).

n(CaI₂) = 12.75 g ÷ 293.9 g/mol.

n(CaI₂) = 0.043 mol; amount of calcium iodide.

In one molecule of calcium iodide, there are two iodine atoms

n(I) = 2 · n(CaI₂).

n(I) = 0.086 mol; amount of iodine atoms.

Na = 6.022·10²³ 1/mol; Avogadro number.

N(I) = n(I) · Na.

N(I) = 0.086 mol · 6.022·10²³ 1/mol.

N(I) = 5.22·10²²; number of iodine atoms.

6 0

3 years ago

Read 2 more answers

Indicar que elemento es de transición *

Musya8 [376]

Answer:

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6 0

2 years ago

At standard temperature and pressure. 0.500 mole of xenon gas occupies

ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

3 0

2 years ago

Be able to identify the correct elements using clues such as numbers of protons,locations (family/period), number of neutrons, n

Brilliant_brown [7]

an element's name, chemical symbol, atomic number, atomic mass.

IDK what you are even asking for

6 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago