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riadik2000 [5.3K]
2 years ago
8

How many grams of water will I obtain from the combustion of 0.9249 mol of pentane (C5H12) in the following reaction?

Chemistry
2 answers:
pshichka [43]2 years ago
8 0

Answer:

C5H12 + 8 (O2) --> 5 (CO2) + 6 (H2O)

*Molar Ratio of Pentane and water is 1:6

*Number of mol of H2O = 0.9249 ×6 = 5.5494 mol

* Molar mass of H2O = 18 g/mol

* Mass of H2O needded = 5.5494 mol × 18 g/mol

= 99.8892 g

Explanation:

* By balancing the equation we can get the stiochemetry of the reaction (Molar Ratio).

* Then we can find the number of mol of H2O neede relative to the no. of mol of Pentane.

* Multiply the mol of H2O by the molar mass of H2O to get the mass.

VLD [36.1K]2 years ago
5 0

Answer: watch?v=zdAtYbCPSnA

Explanation:Should help

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Mr. Jones's prescription calls for 1.04 tablets per day. Based on this information, how many tablets should Mr. Jones take per d
kotykmax [81]

Answer:

c. 1 tablet

Explanation:

1.04 is rounded to 1 so 1 tablet a day is what he needs

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3 years ago
If a doughnut contains 1.8 moles of sugar (C6H12O6), how many sugar molecules are in a box of one dozen doughnuts?
barxatty [35]

Answer:

130.1 x 10²³ molecules of sugar  

Explanation:

One dozen = 12 donuts

1 doughnut = 1.8 moles of sugar

12 doughnut = (1.8 x 12) moles of sugar = 21.6 moles of sugar

1 mole of any substance contains Avogadro’s number (6.023 x 10²³) of molecules.

1 mole of sugar = 6.023 x 10²³ molecules of sugar

21.6 moles of sugar = (6.023 x 10²³ x 21.6) molecules of sugar  

                                 = 130.1 x 10²³ molecules of sugar  

5 0
3 years ago
How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C
yaroslaw [1]

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

  • c= 4.184 \frac{J}{g*C}
  • m=21 g
  • ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 \frac{J}{g*C} * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

<u><em> 527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>

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3 years ago
How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos
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This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

For more information on phosphorylation click on the link below:

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