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AysviL [449]
3 years ago
6

The dissolution of ammonium nitrate occurs spontaneously in water at 25°C. As NH4NO, dissolves, the temperature of the water dec

reases. What are the signs of AH, AS, and AG for this process? a. AH>0, AS > 0, AG > 0 b. AH>0, AS > 0, AG
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

ΔG <0 , ΔH > 0 , ΔS > 0 .

Explanation:

From the data given in question , the reaction is a spontaneous process , hence , the value of change in gibbs free energy would be negative , ΔG <0

And , on dissolution process , the temperature of the water decreases , i.e. , it is an endothermic process , i.e. , the change in enthalphy value is positive , ΔH > 0

And , during the process of dissolution , the ammonia salt break does to ions , i.e. , the randomness increases , hence the ΔS > 0

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The correct answer is letter <span>C. mixture in which its components retain their identity. A heterogeneous mixture is a mixtures in which the component of the mixed are not uniform. You can see that there are localized regions that have different properties. The components have the capacity to retain their identity.</span>
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How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

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Which cell structures does this fish embryo cell contain? Select all that apply.
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Nucleus,mitochondria,& cell membrane

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Light energy travels in the form of
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C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were
melomori [17]

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂  +  2HCl     →     PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number  of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

                            PbCl₂        :          Pb(NO₃)₂

                               1             :             1

                             0.23         :            0.23

Mass of Pb(NO₃)₂:

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Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

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3 years ago
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