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2 years ago

Calculate the mass of hydrogen in 500 g of Al(C2H3O2)3

2 answers:

4 0

22.2 grams For every molecule of Al(C2H3O2)3, we have 9 atoms of hydrogen. So let's first determine how many moles of Al(C2H3O2)3 we have. So start by looking up the atomic weights of the elements involved. Atomic weight aluminium = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass Al(C2H3O2)3 = 26.981539 + (12.0107 * 2 + 1.00794 * 3 + 15.999 * 2)*3 = 204.111199 Moles Al(C2H3O2)3 = 500 g / 204.111199 g/mol = 2.449645107 mol Moles H = 9 * 2.449645107 mol = 22.04680597 mol Mass H = 22.04680597 mol * 1.00794 g/mol = 22.22185761 g Rounding to 3 significant figures gives 22.2 grams.

Zigmanuir [339]2 years ago

3 0

Answer is: the mass of hydrogen is 22,05 grams.
m(Al(C₂H₃O₂)₃) = 500 g.
M(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.
n(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.

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A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.

MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

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Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that

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Answer:

The three-step synthesis of trans-2-pentene from acetylene is as follows.

Step -1: Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.

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Step -3: Formation of trans-pent - 2-pent-ene by reduction.

Explanation:

Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.

The chemical reaction of each step of chemical reactions is as follows.

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BlackZzzverrR [31]

The alignments of the planets would be the correct answer.

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The AE of a system that releases 12.4 J of heat and does 4.2 J of work on its surroundings It is_______ jA. 16.6 B. 12.4 C. 4.2

andre [41]

ANSWER

EXPLANATION

Given that

The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

In the given data, energy is said to be released to the surroundings

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Step 2; Substitute the given data into the formula in step 1

$\begin{gathered} \text{ }\Delta E\text{ = q + w} \\ \text{ }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{ }\Delta E\text{ = -12.4 - 4.2} \\ \text{ }\Delta E\text{ }=\text{ -16.6J} \end{gathered}$

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