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Bess [88]
4 years ago
7

What type of process occurs when solidified wax is converted to liquid wax on heating?

Chemistry
1 answer:
Vika [28.1K]4 years ago
6 0
The physical reaction of melting
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23.1 g of HCl (a strong acid) is added to water to make 1250 mL of solution. Calculate [H3O+] and pH of the solution
MariettaO [177]

Answer:

Following are the solution to this question:

Explanation:

\to HCl \ mol = \frac{23.1 \ g}{36.5 \frac{g}{mol}} = 0.633 \ mole\\\\\\\to HCl \ molarity = \frac{ 0.633 \ mole }{1.25 L} \\\\

                            =0.5064 \ m

 HCl +H_2O \longrightarrow  H_3O^{+} + Cl^{-}\\\\\\

[H_3O^{+}] =[HCl] \\\\\to [H_3O^{+}] =0.5064 \ m \\\\\to PH =-\log [H_3O^{+}] \\\\

          = -\log (0.5064)\\\\= 0.295 \\\\=0.3

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3 years ago
The rate of a chemical reaction is NOT affected by which of the following:
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Answer:

Option B

Explanation:

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Help pls, it's a test about fossils?
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C. At some point, the land was covered in water.

Explanation:

In the diagram it shows the different fossil layers. In the bottom layers there are shells, but as you go up the fossils start changing. This shows that there used to be water there, but now it is land.

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Identify the family to which the compound (h3c−ch2oh) belongs. identify the family to which the compound (a) belongs. alcohol et
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Each of the organic compounds mentioned has a general formula so that we can identify the classification of a certain substance. The compound CH₃CH₂OH is an alcohol because it follows the general formula R-OH, where R is a hydrocarbon chain. In this case, the hydrocarbon chain is ethane. When a hydroxyl functional group is attached, it becomes an alcohol whose name is ethanol.
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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
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