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Usimov [2.4K]
3 years ago
7

The overall reaction in a commercial heat pack can be represented as How much heat is released when 4.40 moles of iron are react

ed with excess ? Heat = kJ How much heat is released when 1.00 mole of is produced? Heat = kJ How much heat is released when 1.60 g iron is reacted with excess ? Heat = kJ How much heat is released when 11.8 g and 1.20 g are reacted? Heat = kJ
Chemistry
1 answer:
vodomira [7]3 years ago
7 0

Answer:

Explanation:

The overall equation for this reaction can be represented as:4 Fe(s) + 3 O_{2(g) }  \to 2Fe_2O_{3(s)} \ \ \ \      \Delta H = -1652  \ kJ

The first question says:

How much heat is released when 4.40 moles of iron is reacted with excess O₂?

Suppose 1652 kJ of heat is being emitted into the surroundings when four(4) moles of Fe reacted with O₂, therefore;

4.40 moles of Fe reacts with:

=\dfrac{4.40 \ moles \times  1652 \ kJ}{4 \ moles}

= 1817.2 kJ of heat will be produced.

The second question says:

How much heat is released when 1.00 mole of Fe_2O_3 is produced?

Given that 1652 kJ of heat is being emitted into the surroundings when two(2) moles of Fe_2O_3 is produced, therefore;

1.00 moles of Fe_2O_3 reacts with:

=\dfrac{1.00 \ moles \times  1652 \ kJ}{2 \ moles}

= 826 kJ of heat will be produced.

To the third question; we have:

How much heat is released when 1.60 g iron is reacted with excess O₂?

We need to find the number of moles of iron first.

We know that number of moles = mass/molar mass

Thus, the molar mass of iron = 55.8 g/mol

number of moles of iron = (1.60g) / (55.8 g/mol)

number of moles of iron = 0.02867 mol

Thus; \dfrac{0.02867\  mol \times  1652 \ kJ }{4 \ mol}

= 11.84 kJ of heat is released.

The last question says:

How much heat is released when 11.8 g Fe and 1.20 g O₂ are reacted?

Again;

the number of moles of Fe = (11.8g) / (55.8 g/mol) = 0.2114 mole of Fe

Thus; \dfrac{0.2114\  mol \times  1652 \ kJ }{4 \ mol}

= 87.31 kJ of heat is released.

On the other hand,

the number of moles of O₂ = (1.20g) / (32 g/mol) = 0.0375 mol of O₂

Thus; \dfrac{0.0375\  mol \times  1652 \ kJ }{3 \ mol}

= 20.65 kJ of heat is released

Therefore, when these two(2) reactants reacted with each other, it is just the smaller amount of heat that would be released because oxygen tends to be the limiting reactant.

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Endocytosis and Exocytosis: Differences and Similarities

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Endocytosis and Exocytosis: Differences and Similarities

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Why is bulk transport important for cells?

Cell membranes are semi-permeable, meaning they allow certain small molecules and ions to passively diffuse through them. Other small molecules are able to make their way into or out of the cell through carrier proteins or channels.

But there are materials that are too large to pass through the cell membrane using these methods. There are times when a cell will need to engulf a bacterium or release a hormone. It is during these instances that bulk transport mechanisms are needed.

Endocytosis and exocytosis are the bulk transport mechanisms used in eukaryotes. As these transport processes require energy, they are known as active transport processes.

Vesicle function in endocytosis and exocytosis

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2 years ago
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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