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olga nikolaevna [1]
2 years ago
7

Look at the diagram of a fuel cell used to power a motorized vehicle. A fuel cell with 2 vertical objects labeled A and B connec

ted by an electrical wire through a circle with a M in it; that circle is labeled A. There is an area between the two vertical objects, and substances flowing to, along, and away from the vertical objects and to the left and right. Which part of the fuel cell does A represent? anode cathode electrolyte motor
Chemistry
2 answers:
77julia77 [94]2 years ago
8 0

Answer:

d

Explanation:

tester [92]2 years ago
4 0

Answer: Its D. motor

Explanation:

TRUST ME ITS RIGHT

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If an organic compound has a carbonyl group and a hydroxyl group, with the carbonyl group connected to two R groups, is it consi
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Answer:

A carboxyl group (COOH) is a functional group consisting of a carbonyl group (C=O) with a hydroxyl group (O-H) attached to the same carbon atom. Carboxyl groups have the formula -C(=O)OH, usually written as -COOH or CO2H.

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When combined in the correct ratio hydrogen and oxygen atoms can form water as shown below
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Answer:molecule of a compound

Explanation:

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Water vapor is heated and rising into the air is called what?
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Answer:

Evaporation

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3 0
3 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
15.0 g of cream at 16.4 °C are added to an insulated cup containing 100.0 g of coffee at 86.5 °C. Calculate the equilibrium temp
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115 grams total

15/115= 13%

109/115= 87%

.13x 16.4= .277

.87x 86.5= 75.255

75.255+.277= 75.532 deg C
or 75.5 as 3 significant digits
7 0
3 years ago
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