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SpyIntel [72]
3 years ago
8

Which type of energy is generated due to the movement of charged particles?

Chemistry
2 answers:
noname [10]3 years ago
7 0
Electrical energy as chare carriers are flowing
____ [38]3 years ago
4 0
Electrical energy or the answer C
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If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need
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In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps
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Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo

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3 years ago
If an atom gains an electron, what overall charge will it have?
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The atom will have a negative charge.

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An equation thats shows how an objects acceleration relates to its mass
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"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

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_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0
3 years ago
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