3 years ago

Explain the sources (SO2 & NOx)

Chemistry

1 answer:

lawyer [7]3 years ago

8 0

Answer:

Explanation:

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worty [1.4K]

C is the answer.
The temperature T<span> in degrees Celsius (°C) is equal to the temperature </span>T<span> in Kelvin (K) minus 273</span>°.

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3 years ago

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steposvetlana [31]

I think It’s 55 but that’s just me

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3 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

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disa [49]

Linear is the answer

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3 years ago

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Leokris [45]

Answer:

Thats all i can think of Hope this helped

Explanation:

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