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3 years ago

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An aqueous solution of sucrose (C12H22O11C12H22O11) is prepared by dissolving 7.6330 gg in sufficient deionized water to form a

25.00 mLmL solution. Calculate the molarity of the solution.

1 answer:

Mariana [72]3 years ago

6 0

Answer:

The answer to your question is 0.64 M

Explanation:

Data

Sucrose C₁₂H₂₂O₁₁ mass = 7.633 g

volume = 25 ml

Molarity = ?

Process

1.- Calculate the molar weight of Sucrose

C₁₂H₂₂O₁₁ = (12 x 12) +(22 x 1) + (11 x 16)

= 144 + 22 + 176

= 342 g

2.- Calculate the moles of sucrose

342 g ------------------ 1 mol

7.633 g --------------- x

x = (7.633 x 1) / 342

x = 0.0223 moles

3.- Calculate the molarity

Molarity = moles / volume (L)

Molarity = 0.0223 / 0.035

Molarity = 0.64

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1 year ago

Magnesium (24.48g) reacts with hydrochloric acid (xg) to produce hydrogen gas (2.04g) and magnesium chloride (96.90g). How much

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Answer:

Mass = 73.73 g

Explanation:

Given data:

Mass of Mg used = 24.48 g

Mass of HCl used = ?

Mass of hydrogen gas produced = 2.04 g

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Solution:

Chemical equation:

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Number of moles of Mg:

Number of moles = mass/ molar mass

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Now we will compare the moles of Mg with HCl from balance chemical equation.

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2 years ago

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Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

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Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

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The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

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The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

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multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

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3 years ago

Substance A and substance B combine to form substance C. Three illustrations of 3-D atomic structures of substances: substance A

ANEK [815]

Answer:

To answer this question we need the illustration

Explanation:

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