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Phantasy [73]
3 years ago
13

What is considered a pure substance?

Chemistry
1 answer:
Fantom [35]3 years ago
4 0

Answer:

A substance that has a fixed chemical composition throughout is called a pure substance such as water, air, and nitrogen. A pure substance does not have to be of a single element or compound.

Explanation:

Compound: A substance that contains more than one element

pure substance: is made up of one substance made up of only one type of molecule

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Samples of three different compounds were analyzed and the masses of each element were determined.
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First, you have to convert mass to moles by making use of molecular weights [N = 14 g/mol and O = 16 g/mol)

A.
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3.2/16 = 0.2 mol O
mol ratio = 2 mol N/1 mol O => N2O

B.
3.5/14 = 0.25 mol N
8.0/16 = 0.5 mol O
mol ratio = 1 mol N/2 mol O => NO2

C.
1.4/14 = 0.1 mol N
4.0/16 = 0.25 mol O
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What is the process by which the movement of sand dunes occurs
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Reactivity and flammability are are examples of properties.​
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Answer:

they are eamples of chemical properties

Explanation:

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You are looking at two samples of iron. One sample is very small and came from the red soil on the planet Mars. The other sample
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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
Read 2 more answers
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