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brilliants [131]
3 years ago
15

A pile of leaves is lit on fire and covered completely with a fire blanket. The fire goes out. Identify the limiting reactant

Chemistry
1 answer:
MAVERICK [17]3 years ago
5 0
Fire blanket i think
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How did changing from a light drizzle to a downpour affect the river and the sediment un it
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Answer:

The velocity of the river increased.

There was more erosion in the stream.

The type of sediment that moved changed.

Explanation:

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Who was the first person to discover the existence of electrons​
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Electron was discovered by J. J. Thomson in 1897 when he was studying the properties of cathode ray.
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3 years ago
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Hot water enters a double-pipe counter-flow water-to-oil heat exchanger at 220°7 and leaves a 100°F. Oil enters at 70°F and leav
il63 [147K]

Answer:

Oil has the smaller heat capacity. The effectiveness of the heat exchanger is 0.80.

Explanation:

Part 1:

In order to know which fluid has the smaller heat capacity we need to consider the heat equation below:

Q = CΔT, where Q is the heat exchanged, C is the heat capacity and ΔT is the variation in temperature.

As the heat exchange is the same for both fluids, the smaller the temperature variation, the smaller the heat capacity.

Water: ΔT = 120 °F  

Oil: ΔT = 80 °F  

Therefore, oil is the fluid with the smallest heat capacity.

Part 2:

The effectiveness of a counter-flow heat exchanger is given by the equation bellow:

E = \frac{Th1 - Th2}{Th1 - Tc1} \\

Th1: initial temperature of the hot fluid

Th2: final temperature of the hot

Tc1: initial temperature of the cold fluid

E = \frac{220 - 100}{220 - 70} \\E = 0.8

7 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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