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Novosadov [1.4K]
2 years ago
14

Moist to wet as warm is to: Luke Boiling or Steamy

Chemistry
1 answer:
Vinvika [58]2 years ago
8 0

Answer:steamy

Explanation:

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When sand is coated with a layer of trimethylhydroxysilane, (CH3)3SiOH, it repels water and can no longer get wet. Hydrophobic s
Juli2301 [7.4K]
Is that how it’s spell trimethyhydrosilane?
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2 years ago
When sodium reacts with chlorine, sodium chloride is produced. Andrew represented this reaction with this equation: Na + Cl2 → 2
My name is Ann [436]
When sodium metal reacts with chlorine gas, the product would be sodium chloride or the table salt. The balanced chemical reaction would be:

2Na + Cl2 = 2NaCl

IN balancing reactions, it is important to remember that the number of atoms at each side should be equal. Hope this answers the question.
4 0
2 years ago
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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
ira [324]

Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

6 0
2 years ago
2.1 liters is the same as cm3 and ml
Kipish [7]
1L = 1000ml
1ml = 1cm^3

2.1L = 2100mL = 2100cm^3

4 0
2 years ago
Jim takes 45 seconds to walk 180 meters north to a store what is jims meters per second
Delicious77 [7]
It would be 4 meters per second. With this you'd only have to take 180, ans divide 45 from it to finally get your answer! I hope all is well, and you end up passing. (:
7 0
2 years ago
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