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2 years ago

Describe a method to investigate how the temperature changes when different masses of ammonium

Chemistry

1 answer:

rewona [7]2 years ago

4 0

Answer:

Decrease temperature with increase in ammonium nitrate concentration.

Explanation:

The temperature changes when different masses of ammonium nitrate are dissolved in water because the ammonium nitrate takes energy from the surrounding water environment for the breaking of ionic bonds as a result the temperature of water decreases. When more concentration of ammonium nitrate is added to water, more decrease occur in temperature of water and the water becomes cold.

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Alai is comparing the physical property of two materials. He is hitting each with a hammer to observe what happens. What physica

maxonik [38]

Since Alai is hitting each with a hammer, the physical property which he must be comparing on the two materials must be hardness. Solubility can be tested when you put them in a solvent. Odor can be tested with smell. Conductivity can be tested with electricity. Hardness can be tested with its resistance to the force that hits it.

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3 years ago

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Given the formula: what is a chemical name of this compound? propane propanal propanol propanone

katrin [286]

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3 years ago

When magnesium reacts with hydrochloric acid, hydrogen gas is produced. The amount of gas produced is dependent upon the concent

velikii [3]

Answer:

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1 year ago

Whuch expression is equal to the number of grams in 2.43 kg

schepotkina [342]

2.43*1000.

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3 years ago

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

= 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0

3 years ago