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Natasha2012 [34]
3 years ago
9

The group in an experiment that is not exposed to the tested variable is called the group

Chemistry
1 answer:
kati45 [8]3 years ago
5 0

Answer:

yes your answer is correct for this question.

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An atom of titanium has 22 protons and 26 neutrons. If the atoms loses one electron, what will be the charge on the ion that for
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It’s +21 (c) if i answered late my apologies
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2 years ago
5TH GRADE SCIENCE!!!!!!
Yuri [45]

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Phase changes that require a loss in energy are condensation and freezing.

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4 0
3 years ago
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Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
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I need help please my teaching is going to go crazy if I didn’t do it
victus00 [196]

Answer:

No One is going to answer all this because it hard to type it all in her. its easy to search it up

Explanation:

4 0
3 years ago
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