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Anastaziya [24]
3 years ago
11

Matthew records the temperature outside for one day. From 10:00 a.m. to 12:00 p.m., the

Chemistry
1 answer:
ratelena [41]3 years ago
7 0

Answer:

The temperature will drop 4°C.

Explanation:

Note that this question is describing temperature change in response to the time of the day. Ideally, mornings to afternoons are usually hotter while evenings are colder.

According to this question, the temperature rises by 4°C from 10:00 a.m. to 12:00 p.m. it also rises another 3°C from 2:00 p.m. to 4:00 p.m but starts dropping by 2°C from 2:00pm to 4:00pm. This shows that between the hours of 10:00am - 2:00pm, the temperature increases as a result of the heat from sun. However, towards the evening period, the temperature reduces because the sun is getting shadowed by the moon.

Hence, based on this analogy, the most reasonable prediction about how the temperature will change from. 4:00am to 6:00am is that The temperature will drop 4°C.

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Molecules can be either positively or negatively charged depending upon their elemental arrangement
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They are positively or negatively charged based on their electrical configuration of electrons*
For example an electronic configuration of 2,8,3 would have a negative charge if +3 since it needs to lose 3 electrons to gain the electrical configuration of a noble gas
2,8,1 would have a charge of +1 for the same reason
2,8,6 would be -2 since it is easier to gain 2 electrons that lose 6 electrons

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3 years ago
Help for this practice work
allochka39001 [22]

Answer:

It's the third option.

Explanation:

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2 years ago
The smith family uses more energy in the summer than the winter what is a good explanation for this?
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3 years ago
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Q7-A graduated cylinder is filled to the 12.0 mL line with water. A solid with a mass of 14.52 g is placed in the graduated cyli
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<h3><u>c) 13.29 mL</u></h3>

6 0
2 years ago
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
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