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vampirchik [111]
2 years ago
8

At 850K, 65L of gas has a pressure of 450kPa. What is the volume (in liters) if the gas is cooled to 430K and the pressure decre

ases to 325kPa?
Chemistry
1 answer:
Novosadov [1.4K]2 years ago
8 0

Answer:

V₂ =  45.53 L

Explanation:

Given data:

Initial temperature = 850 K

Initial volume = 65 L

Initial pressure = 450 KPa

Final temperature = 430 K

Final pressure = 325 KPa

Final volume = ?

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa  

V₂ = 12577500 KPa .L. K / 276250 K. KPa

V₂ =  45.53 L

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How many iodide ions are present in 65.5ml of .210 m AlI3 solution
blondinia [14]

Answer:

2.48\times 10^{22} ions are present in solution.

Explanation:

Molarity of the solution = 0.210 M

Volume of the solution = 65.5 ml = 0.0655 L

Moles of aluminum iodide= n

Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}

0.210M=\frac{n}{0.0655 L}

n = 0.013755 moles of aluminum iodide

1 mole of aluminum iodide contains 3 moles of iodide ions:

Then 0.013755 moles of aluminum iodide will contain:

3\times 0.013755 moles=0.041265 mol of iodide ions

Number of iodide ions in 0.041265 moles:

0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions

2.48\times 10^{22} ions are present in solution.

7 0
2 years ago
A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?
Dominik [7]

Answer:

The specific heat of the metal is 0.466 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

  • Q= 2330 J
  • c= ?
  • m= 25 g
  • ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

c=\frac{2330 J}{25 g* 200 C}

c=0.466 \frac{J}{g*C}

<u><em>The specific heat of the metal is 0.466 </em></u>\frac{J}{g*C}<u><em></em></u>

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2 years ago
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Answer:

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Explanation:

Hello,

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