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Studentka2010 [4]
3 years ago
6

A gas occupies 900.0 mL at a temperature of 27.0 . What is the volume at 132.0 °C?

Chemistry
1 answer:
Ronch [10]3 years ago
7 0
V₁ = 900.0 mL

T₁ = 27.0ºC + 273 = 300 K

V₂ = ?

T₂ = 132.0 + 273 = 405 K

V₁ / T₁ = V₂ / T₂

900.0 / 300 = V₂ / 405

300 x V₂ = 405 x 900.0

300 x V₂ = 364500

V₂ = 364500 / 300

V₂ = 1215 mL

hope this helps!
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V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

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

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

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7 0
2 years ago
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klemol [59]

Answer:

a) 2.01 g

Explanation:

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First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

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So the remaining Na₂CO₃ moles are:

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Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

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The closest answer is option a).

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3 years ago
How many kilojoules are released when 2.25 mol of H2O2 reacts?
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Answer:

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Using dielectric constant for Air=8.84×10-12F/m

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Answer:

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Explanation:

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