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Serga [27]
3 years ago
5

How many square centimeters are in an area of 9.90 inch2 ?

Chemistry
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

63.8708 hope this helps :)

Explanation:

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Describe how glacial erosion influenced the topography of Ohio.
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During the past two million years, glaciers have shaped and reshaped the surface of Ohio several times. These continental masses of ice affected as much as two-thirds of the state. Moving from the north and northwest, glaciers have scraped and flattened the landscape. Often more than a mile thick, they smoothed existing hills and filled valleys with enormous amounts of rocks, gravel, and smaller particles.  Through these actions, glaciers have had a very important impact on the agriculture of Ohio. Their activity has been felt in two noticeable ways: shaping the ground upon which people work and build, and forming the soils that cover that ground.

4 0
3 years ago
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.


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3 years ago
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Hello!

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2) Dissociation of the salt of a weak acid:

NH₄I → NH₄⁺(aq) + I⁻ (aq)

3) Dissociation of a weak acid to form H₃O⁺ (very little):

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

This series of reactions show how the adding of a strong Acid can be neutralized by the buffer, releasing instead very little amounts of Hydronium ions.

Have a nice day!
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