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leva [86]
3 years ago
13

Atoms become excited when they are heated by a flame. How do atoms return from an excited state to a ground state?

Chemistry
1 answer:
ohaa [14]3 years ago
8 0
Atoms return to their ground state by emitting a photon of light.
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What is the limiting factor in determining the accumulation of siliceous ooze/calcareous ooze, respectively?
zhannawk [14.2K]

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.  

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

8 0
3 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
What's an element in period 5 with larger radius than Yttrium?
eimsori [14]
Rubidium or strontium have larger a larger atomic radius since the further left on the periodic table you go, the larger the sizes of the atoms are.  This trend can be explained through effective nuclear charge which explains how the further left and down you go, the less the atoms nucleus is able to pull in the electrons around it.<span />
7 0
3 years ago
Find the % by mass of the solute in 129.54 g C6H1206 in 525 g H20.
belka [17]

Answer:

19.79%

Explanation:

mass % = (mass solute / total mass) * 100

total mass = 129.54 + 525 = 654.54

solute = C6H12O6

(129.54/654.54) = .1979

.1979 * 100 = 19.79%

7 0
3 years ago
Adolfo drank some red punch and grimaced at the flavor. His little brother had made the pitcher of punch with too much sugar. Wh
drek231 [11]
I believe that the saliva interacting with sugar molecules is what allowed Adolfo’s brain to register the extra-sugary flavor of the punch
7 0
3 years ago
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