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stiks02 [169]
2 years ago
15

List the following objects in order from lowest inertia to highest inertia

Chemistry
2 answers:
m_a_m_a [10]2 years ago
5 0
I don’t see nun tho where’s the objects
kumpel [21]2 years ago
5 0

Earth, desk,baseball, fork, house, feather, tow truck, cow, sun

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Desertification.......
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A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate
kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of solution 1 (liquid water) = 50.0 g

m_2 = mass of solution 2 (liquid water) = 29.0 g

T_{final} = final temperature = ?

T_1 = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

T_2 = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K

Converting this into degree Celsius, we use the conversion factor:

T(K)=T(^oC)+273

305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC

Hence, the final temperature of water is 32.3°C

7 0
3 years ago
What part of the sun do we see from earth?
NISA [10]

Answer:

photosphere

Explanation:

photosphere

There are 3 main layers of the Sun that we can see. They are the photosphere, the chromosphere and the corona. Together they make up the "atmosphere" of the Sun. The part of the Sun that glows (and that we see with the naked eye) is called the photosphere

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3 years ago
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Which element has atoms with only one completely filled principal energy level
liubo4ka [24]
I think it’s hydrogen
3 0
3 years ago
An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down
jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0
3 years ago
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