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kirill [66]
3 years ago
12

A higher concentration of molecules causes a faster chemical reaction this is known as

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
3 0
Chemical equilibrium is the answer you're looking for.

Hope this proves helpful to you!
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Match each energy transformation to the correct image. ANSWER FAST PLS!!! THIS IS FOR 20 POINTS!!! AND I WILL MARK BRAINLIEST IF
lord [1]

Answer:

Explanation:

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Draw the skeletal structure for the monohalogenated alkene formed when trans-1-chloro-4-iodocyclohexane is reacted with 1 equiva
Ksivusya [100]
Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />

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3 years ago
How many moles are in 17.2g k2s
Natalija [7]

Answer: 0.156 mol

Explanation:

To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.

17.2g*\frac{mol}{110.256 g} =0.156 mol

3 0
3 years ago
Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
zysi [14]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

N_2+2O_2\rightarrow 2NO_2

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0
3 years ago
Convert 32 Fahrenheit to Celsius
valina [46]

Answer:

°F - 32/1.8000

Explanation:

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