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3 years ago

In a graph, what effect would adding a catalyst have on D?

Chemistry

2 answers:

Anika [276]3 years ago

7 0

It would decrease hope this helps

CaHeK987 [17]3 years ago

3 0

Answer:

C. It would stay the same

Explanation:

PLATO answer

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Stained glass consists of colored dyes mixed with silicon dioxide molecules. Variations in the color of the glass occur because

Juli2301 [7.4K]

Answer:

Explanation:

heterogeneous!!!!

6 0

3 years ago

For the reaction

iVinArrow [24]

Answer:

16 mol NaCl.

Explanation:

Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.

179.2 L CO2 x 1 mol CO2/22.4 L CO2 x 2 mol NaCl/1 mol CO2

= 16 mol NaCl

5 0

2 years ago

7.20g of potassium sulfate reacts with the excess oxygen gas in a single displacement reaction, how many grams of oxygen are con

skad [1K]

The balanced chemical reaction:

K2SO4 + O2 = 2KO2 + SO2

Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:

7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.

5 0

3 years ago

NEED ASAP

schepotkina [342]

Non of these bc it makes the most likely

5 0

3 years ago

Read 2 more answers

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

5 0

3 years ago