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lutik1710 [3]
3 years ago
12

If m< CAB = 80 degrees, then find m< BAD

Chemistry
1 answer:
eimsori [14]3 years ago
7 0

Answer:

30

Explanation:

Just trust me

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Urgent please help me!!!
nikitadnepr [17]

Explanation:

A. lithium hydroxide =LiOH

B. sodium cyanide =NaCN

C. Magnesium nitrate = Mg(NO3)2

D. Barium sulfate = BaSO4

E. Aluminum nitride = AlN

F. Potassium phosphate = KH2PO4

G. Ammonium bromide = NH4Br

H.Calcium carbonate = CaCO3

5 0
3 years ago
A very loud sound has a high _____.
DENIUS [597]

Answer:

Resting Point

Explanation:

4 0
2 years ago
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Select the correct answer.
rusak2 [61]
I think the answer to this A but I’m not rlly sure
6 0
3 years ago
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How many rods will equal 247 inches
d1i1m1o1n [39]
It will be 1.247 rods that will equal 247 inches.
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3 years ago
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An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
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