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harina [27]
3 years ago
9

Given the reaction:

Chemistry
1 answer:
lisov135 [29]3 years ago
5 0
H20() and NH4+(aqf)
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Why are there more fossils of marine organism than of land organisms o dc
MatroZZZ [7]

Answer:

Three primary reasons. First, there is simply more water-covered places than dry ground places for the animals and plants to have lived. Second, the seas are much more crowded with the kinds of life that leave fossils than the land is. Third, the process that form fossils work very well under water.

Explanation:

4 0
3 years ago
g Suppose the formation of nitrosyl chloride proceeds by the following mechanism: step elementary reaction rate constant That is
PIT_PIT [208]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

5 0
2 years ago
Are the charges in Li2O balanced yes o No?
motikmotik

Answer: For example, the Li 2 O content of the principal lithium mineral spodumene (LiAlSi 2 O 6) is 8.03%. Burning lithium metal produces lithium oxide. Lithium oxide is produced by the thermal decomposition of lithium peroxide at 300-400°C.

3 0
2 years ago
HELP ME NOW ASAP PLEASE, 10 POINTS HURRY
xxMikexx [17]
Which of those depend on mayfly nymphs. In other terms which animal eats mayfly nymphs?

I would say frogs
7 0
2 years ago
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
3 years ago
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