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kotykmax [81]
2 years ago
6

10 points. Please help.

Chemistry
1 answer:
sergiy2304 [10]2 years ago
3 0

Answer:

The temperature of the balloon is 46.34 K.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 0.76 atm
  • V= 10 L
  • n= 2 moles
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= ?

Replacing:

0.76 atm* 10 L= 2 moles* 0.082 \frac{atm*L}{mol*K} *T

Solving:

T=\frac{0.76 atm* 10 L}{2 moles* 0.082\frac{atm*L}{mol*K}}

T= 46.34 K

<u><em>The temperature of the balloon is 46.34 K.</em></u>

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Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

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I hope it helps!

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3 years ago
Meteorologists use weather balloons to carry weather instruments high into the atmosphere. When it is first released at Earth’s
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The pressure of the gas used in the weather balloon increases to expand the balloon.

Explanation:

  • Weather balloons contain the boxes where the weather measurement instruments are present that is attached to the large balloon.
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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
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Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

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at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

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∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

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⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

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