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Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH4NO3(s)→NH+4(aq)+NO−3(aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 ∘C and the final temperature (after the solid dissolves) is 21.9 ∘C. Part A Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/g⋅∘C as the specific heat capacity.) Express your answer to two significant figures and include the appropriate units. ΔHrxn = ??? kJ/mol
Answer:
26 kJ / mol
Explanation:
Given that;
The mass of NH₄NO₃ = 1.25 g
Number of moles of NH₄NO₃ = Mass of NH₄NO₃ / Molar Mass of NH₄NO₃
Number of moles of NH₄NO₃= 1.25 g / 80.043 g/mol
Number of moles of NH₄NO₃= 0.016 mol
Volume of solution = 25.0 mL
Density of Solution = 1.0g/mL
Since;
Mass of Solution = Density x Volume
= 1.0g/mL × 25.0mL
= 25 g
Heat Generated (Q) =
Q= 25g × 4.18 J/g°C x (25.8°C - 21.9°C)
Q = 407.55 J
Q = 407 × 10 ⁻³ kJ
Q = 0.40755 kJ
Δ =
=
= 25.47 kJ/ mol
~ 26 kJ / mol
Therefore, the change in enthalpy for the reaction in kilojoules per mole = 26 kJ / mol
Answer:
B. NaBr
D. KOH
Explanation:
Below is the solubility rules given for you knowledge.
Salts of
- Group 1 elements are soluble(
)
- Ammonium ion is soluble (
)
- The nitrate are generally soluble(
- of Cl- , Br- , and I- are soluble, except Ag+ , Pb+2, and (Hg2)+2
- most sulfate are soluble, except Ba+2, Ca+2,Pb+2, Ag+, Sr+2.
- most hydroxide salts are only slightly soluble, except NH+4, Li+, Na+, K+
- Most carbonates are insoluble (CO3 2-) Except group 1 and NH+4
- most phosphate are insoluble except group 1 and NH+4
so using the rules above
NaBr , KOH are soluble, Pb(OH)2 is slightly soluble and AgCl is not soluble.