Which atmospheric gas molecules were not affected by the visible orinfrared radiation?

1 answer:

8 0

For example, nitrogen (N2) and oxygen (O2), which make up more than 90% of Earth's atmosphere, do not absorb infrared photons.

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BRAINLIST!!

Natali5045456 [20]

B erosion is the answer

6 0

3 years ago

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A sample of water with a mass of 27.56g and an unknown temperature loses 2443 Joules. If the final temperature is found to be 62

mixer [17]

Answer:

41.3 °C

Explanation:

From the question given above, the following data were obtained:

Mass (M) of water = 27.56 g

Heat (Q) loss = 2443 J

Final temperature (T2) = 62.5 °C

Initial temperature (T1) =?

NOTE: The specific heat capacity (C) of water is 4.18 J/g°C

Thus, we can obtain the initial temperature of the water by using the following formula:

Q = MC(T2 – T1)

2443 = 27.56 × 4.18 (62.5 – T1)

2443 = 115.2008 (62.5 – T1)

Divide both side by 115.2008

2443 / 115.2008 = (62.5 – T1)

21.20645 = 62.5 – T1

Collect like terms

21.20645 – 62.5 = – T1

– 41.3 = – T1

Divide both side by – 1

– 41.3 /– 1= – T1 / –1

41.3 = T1

T1 = 41.3 °C

Thus, the initial temperature of the water was 41.3 °C

8 0

3 years ago

How much heat is absorbed by a 112.5 g sample of water when it is heated from 12.5 °C to 92.1 °C? (Specific heat capacity of wat

Alborosie

Answer:

$\boxed {\boxed {\sf37,467.72 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of water absorbed. Since we are given the mass, temperature, and specific heat, we will use the following formula.

$q=mc \Delta T$

The mass (m) of the sample is 112.5 grams. The specific heat capacity of water (c) is 4.184 Joules per gram degree Celsius. The difference in temperature (ΔT) is found by subtracting the initial temperature from the final temperature.