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LenaWriter [7]
3 years ago
6

80.0 mL of oxygen gas was collected when the temperature was

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

89.3 mL

Explanation:

First we <u>convert 20.0 °C and 54.0 °C to K</u>:

  • 20.0 °C + 273.16 = 293.16 K
  • 54.0 °C + 273.16 = 327.16 K

With the absolute temperatures we can solve this problem by using <em>Charles' law</em>:

  • T₁V₂=T₂V₁

Where in this case:

  • T₁ = 293.16 K
  • V₂ = ?
  • T₂ = 327.16 K
  • V₁ = 80.0 mL

We <u>input the data</u>:

  • 293.16 K * V₂ = 327.16 K * 80.0 mL

And <u>solve for V₂</u>:

  • V₂ = 89.3 mL
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Answer : Colloid

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3 years ago
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If you initially have a volume of 4 L and a temperature of 300K. Then you decreased the volume to 2 L, what is the new temperatu
ASHA 777 [7]

Answer:

T₂ = 150 K

Explanation:

Given data:

Initial volume = 4 L

Initial temperature = 300 K

Final volume = 2 L

Final temperature = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 300 K × 2L / 4 L

T₂ = 600 L.K / 4 L

T₂ = 150 K

3 0
3 years ago
Elements that share properties of both metals and nonmetals?
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<span>-Boron (B)
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-Germanium (Ge)
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-Antimony (Sb)
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Which statement about a chemical reaction is true?
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A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
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Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

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