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Kruka [31]
3 years ago
15

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

55 degrees C is mixed with the water in the insulated container. When thermal equilibrium is reached, the temperature of the mixture is 25 degrees C. Assume that heat flows only between the alloy and the water. What is the specific heat of the alloy?
Chemistry
1 answer:
Lorico [155]3 years ago
3 0

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

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If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?
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If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

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pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

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We will calculate the concentration of OH⁻ by using Ksp expression

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Calculate the pOH

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You can also learn about molarity from the following question:

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7 0
1 year ago
I need help with chemistry :(
Yanka [14]

Answer:

How may we help kind sir

Explanation:

and if this was for points thanks

4 0
3 years ago
Read 2 more answers
ASAP worth 15 points also
Taya2010 [7]
The person above me is correct I took a test on this so it’s the right answer
7 0
3 years ago
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