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Katena32 [7]
2 years ago
14

Identify the equations that show ionization. Check all that apply.

Chemistry
1 answer:
tigry1 [53]2 years ago
8 0

Answer: B. HCl(g)+H2O(I)—>H3O+(aq)+Cl-(aq)

D. CO2(g)+2H2O(I)—>HCO3-(aq)+H3O+(aq)

Explanation: on edge

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C
oksano4ka [1.4K]

Answer:

a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.

b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.

The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.

c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.

Explanation:

8 0
2 years ago
What characterizes an Arrhenius acid
dedylja [7]
The characteristics of an Arrhenius acid was the acid donates H+ when it is in the aqueous solution, whether the solution is acid or base. If the solution is a acid, then be H+ builds up and increase its thing. The pH level for acid solution is the scale from 1-6. Even though the ph scale is lower than the base, the color also varies. It also applies to the base, too.
8 0
2 years ago
Nicole has 2 glasses on the counter: one of water and one of sugar. She is baking a cake and needs to use the sugar-water for th
Schach [20]

Answer:It is a mixture

Explanation:

If it is pure sugar it’s neither but if it has water it is a homogeneous mixture

8 0
1 year ago
Read 2 more answers
What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercu
Olegator [25]

Answer:

\Delta T=194^oC

Explanation:

Hello,

In this case, in terms of the heat, mass, heat capacity and change in temperature, we can analyze thermal changes as:

Q=mCp\Delta T\\

In such a way, we compute the required change in temperature as shown below:

\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033\frac{cal}{g^oC} }  \\\\\Delta T=194^oC

Such change in temperature is positive indicating an increase in the temperature as the involved heat is positive, in means that heat was added to increase the temperature.

Best regards.

6 0
1 year ago
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