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tekilochka [14]
3 years ago
5

When 5.00 g of NH4NO3 is dissolved in 100.0 g of water in a styrofoam cup, the T = - 3.82 ^°C. Is the process of dissolving ammo

nium nitrate endo- or exothermic? What is the heat of this process in kJ/mol?
Chemistry
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

The process is endothermic.

Heat is 26,9 kJ/mol

Explanation:

The dissolution of NH₄NO₃ in water is:

NH₄NO₃ → NH₄⁺ + NO₃⁻

As the change of temperature in the cup is ΔT = -3,82°C

<em>The process is endothermic</em>. Because the temperature is decreasing in the process. That means the process needs heat.

Assuming the  heat capacity of the solution is 4.18 J/Kg :

q = -C×m×ΔT

Where q is heat, C is heat capacity (4,18J/Kg), m is mass (105g) and ΔT is change in temperature (-3,82°C)

q = 1677 J ≈ <em><u>1,68kJ</u></em>

The moles of NH₄NO₃ dissolved are:

5,00g × (1mol / 80,043g) = <em><u>0,0625 moles</u></em>

That means heat of this process in kJ/ mol is:

1,68kJ / 0,0625moles = <em>26,9 kJ/mol</em>

<em />

I hope it helps!

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Read 2 more answers
A gas is collected in a 34.3L container at a temperature of 31.5°C. Later, the container has a volume of 29.2L, a temperature of
goldenfox [79]

Answer:

108 kPa  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂             Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂      Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

Data:

p₁ = ?;                 V₁ = 34.3 L; T₁ = 31.5 °C

p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C

Calculations:

(a) Convert temperatures to <em>kelvins </em>

T₁ = (31.5 + 273.15) K = 304.65 K

T₂ = (21.0 + 273.15) K = 294.15 K

(b) Calculate the <em>pressure </em>

p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)  

   = 122.2 kPa × 0.8542 × 1.0357

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4 0
3 years ago
Calculate the molarity of 90.0 mL of a solution that is 0.92 % by mass NaCl. Assume the density of the solution is the same as p
jeka94
Answer is: molarity is 0,155 M.
V(solution) = 90,0 mL = 0,09 L.
ω(NaCl) = 0,92% ÷ 100% = 0,0092.
d(solution) = 1 g/mL.
m(solution) = V(solution) · d(solution).
m(solution) = 90 mL · 1 g/mL = 90 g.
m(NaCl) = 90 g · 0,0092 = 0,828 g.
n(NaCl) = 0,828 g ÷ 58,4 g/mol.
n(NaCl) = 0,014 mol.
c(solution) = 0,014 mol ÷ 0,09 L.
c(solution) = 0,155 mol/L.

8 0
3 years ago
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