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Xelga [282]
3 years ago
10

I need an example of a chemical change involving water.Pleaseeee.

Chemistry
1 answer:
slava [35]3 years ago
7 0
Chemical change occur when two substances are combined and produces a new substance or decomposes into two or more substances which are entirely different from the original two substances.

There are three types of chemical changes. These are 1) Inorganic Changes, 2) Organic Changes, and 3) Biochemical Changes


Here are some examples of chemicsal changes.

If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.

Sodium + Water  ==>  Sodium Hydroxide and Hydrogen

Na + H2O ====>  NaOH and H

Another example of chemical change is:

Carbon Dioxide and Water will decompose into Sugar and Oxygen

Carbon Dioxide + Water ==> Sugar and Oxygen

CO2 + H2O ==> CnH2nOn (where n is between 3 and 7) and O

 
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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
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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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