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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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Haha those three make up a simple Atom.
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Determine the chemical formulas for the two compounds below. (Carbon, Hydrogen, and Oxygen).
tester [92]
Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.

1) CH3 - CH (OH) - CH (CH3) -CH3

2) CH3 - O - CH(CH3)-CH2 - CH3

I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -
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3 years ago
What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
My name is Ann [436]

Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
3 years ago
Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution
Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0
3 years ago
Concentrations-
GaryK [48]
Final molarity is 0.607M

Work shown on photo:

3 0
3 years ago
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