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Hunter-Best [27]
3 years ago
5

What is the enthalpy of fusion of water?

Chemistry
1 answer:
malfutka [58]3 years ago
5 0

Answer:

d-  334 kJ/g.

Explanation:

You can detect it from the units of the different choices.

a- has the unit J/g.°C that is the unit of the specific heat capacity (c).

b- has the unit Kelvin that is the unit of temperature.

c- has the unit g/mol which is the unit of the molar mass.

d- has the unit kJ/g which is the unit of the enthalpy divided by the no. of rams that is the specific entha;py of fusion.

<em>So, the right choice is: d-  334 kJ/g.</em>

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In plant cells, how are chloroplasts and mitochondria related?
guapka [62]

Answer:

Mitochondria are the "powerhouses" of the cell, breaking down fuel molecules and capturing energy in cellular respiration. Chloroplasts are found in plants and algae. They're responsible for capturing light energy to make sugars in photosynthesis.

5 0
3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

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3 years ago
Determine the mass of 4.20 moles of C6H12
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Explanation:

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A Bronsted-Lowry<br> _______is a molecule or ion that donates<br> a hydrogen ion in a reaction,
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A Bronsted-Lowry acid-base is a molecule or ion that donates a hydrogen ion in a reaction.

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Answer:

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