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lana66690 [7]
3 years ago
14

Calculate the mass of H2OH2O produced by metabolism of 1.6 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11

0O6C57H110O6), a typical animal fat, and assuming that during metabolism, tristearin reacts with O2O2 to form only CO2CO2 and H2OH2O.
Chemistry
1 answer:
kakasveta [241]3 years ago
5 0

Answer: 1779.8g of H2O

Explanation:

C57H110O6 + O2 —> CO2 + H20

To calculate the amount of water(H2O) produced by the reaction of 1.6Kg tristearin(C57H110O6) and O2, we must first balance the equation.

To do so, add 2 in front of C57H110O6, 163 in front of O2, 114 in front of CO2 and 110 in front of H2O i.e

2C57H110O6 + 163O2 —> 114CO2 + 110H20

Next, we have to calculate the mass of tristearin(C57H110O6) that reacted and the mass of water produced from the balanced equation. This can be done by:

MM of C57H110O6 = (57x12) + (110x1) + (6x16) = 684 + 110 + 96 = 890g/mol

Mass conc of C57H110O6 = 890 x 2 = 1780g

MM of H2O = (2x1) + 16 = 18g/mol

Mass conc of H2O = 110 x 18 = 1980g

Next, we will find the amount of water produced by 1.6Kg(i.e 1600g) of tristearin(C57H110O6). This is done by:

From the equation,

1780g of C57H110O6 produced 1980g of H2O.

Therefore 1600g of C57H110O6 will produce = (1600x1980)/1780 = 1779.8g of H2O

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Zinc Sulfide reacts with oxygen according to the reaction:
zhuklara [117]

The amount in moles of the excess reactant left is 0.5 mole

<h3>Balanced equation</h3>

2ZnS (s) + 3O₂(g) --> 2ZnO (s) + 2SO₂(g)

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

<h3>How to determine the excess reactant</h3>

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

Therefore,

4.2 moles of ZnS will react with =(4.2 × 3) / 2 = 6.3 moles of O₂

From the calculations made above, we can see that only 6.3 moles of O₂ out of 6.8 moles given, is required to react completely with 4.2 moles of ZnS.

Thus, ZnS is the limiting reactant and O₂ is the excess reactant.

<h3> How to determine the mole of the excess reactant remaining</h3>

The excess reactant is O₂. Thus the mole remaining after the reaction can be obtained as illustrated below:

  • Mole of O₂ given = 6.8 moles
  • Mole of O₂ that reacted = 6.3 moles
  • Mole of O₂ remaining =?

Mole of O₂ remaining = (Mole of O₂ given) - (Mole of O₂ that reacted)

Mole of O₂ remaining = 6.8 - 6.3

Mole of O₂ remaining = 0.5 mole

Learn more about stoichiometry:

brainly.com/question/25685654

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6 0
2 years ago
Pleaseeee help me only 29 and 30
stepladder [879]

Answer:

29: a. unbalanced

b. unbalanced

c. unbalanced

30. a. coefficient b. subscript c. formula unit d. element e. reactant f. product

3 0
3 years ago
A potassium permanganate solution is prepared by dissolving 22.14g of KMnO4 in water and diluting to a total volume of 1.000 L.
gladu [14]

Answer:

The concentration of H₂C₂O₄ is 7,1800x10⁻³ M

Explanation:

The reaction of the titration is:

2 KMnO₄ + H₂C₂O₄ → 2 CO₂ + K₂O + 2 MnO₃ + H₂O

22.14g of KMnO₄ in 1,00L of water have a molarity of:

22,14g KMnO₄/L×\frac{1mol}{158,034g} = 0,14 M of KMnO₄

The moles you required for a complete reaction are:

0,14 mol KMnO₄/L× 0,02050L = 2,872x10⁻³ mol KMnO₄

By the reaction 2 moles of KMnO₄ reacts with 1 mole of H₂C₂O₄, thus, the moles of H₂C₂O₄ that react were:

2,872x10⁻³ mol KMnO₄×\frac{1mol H_{2}C_{2}O_{4}}{2molKMnO_{4}} = 1,436x10⁻³ moles of H₂C₂O₄

As the volume of the sample was 200,0mL≡ 0,2000L, The concentration of H₂C₂O₄ is:

\frac{1,436x10^{-3} mol}{0,2000L} = <em>7,1800x10⁻³ M</em>

<em />

I hope it helps!

4 0
4 years ago
52. In the following reaction, which are the Spectator ions? *
maks197457 [2]

Answer:

nitrate and sodium are spectator ions .The reactions that form an aqueous solution are often a spectator ions.

5 0
2 years ago
Solution 1 contains 3.5g of sodium chloride in 500 mL of water. Solution 2 contains 7.5g of glucose in 300 mL of water. If the a
meriva

Answer:

Answer is Object 2 (which has a density of 1.9 g/cm³).

Explanation;

   When object is floating, the weight of that object is less than the up thrust on it.

   When an object fully submerged and floating, then the weight of that object is equal to the up thrust on it.

   This is known as the Archemide's principle.

   Both up thrust and weight depends on the density. Hence, if the density of the solution is high, then the up thrust also high. If the density high, the the weight of the object also high.

   Hence, to sink the object in water, that object should be denser than water. Hence, answer is object 3 which has a higher density than water.

Explanation:

8 0
3 years ago
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