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3 years ago

14

Your teacher directs you to combine two liquid substances into a flask. after you combine them, you notice that the flask feels

cold when you touch it. which if the following best depicts what happened?
A: an endothermic reaction has taken place

B: an exothermic reaction has taken place

C: the liquid has reached its freezing point

D: you have made an error because the flask should have gotten warmer, not colder

1 answer:

adelina 88 [10]3 years ago

3 0

Answer : The correct option is (A).

Explanation :

Endothermic reaction : When two liquids are combine into a flask, then the flask feels cold when we touch it because the system absorbed heat from the surrounding.

In general, endothermic process absorbs heat and cool the surrounding.

Exothermic reaction : When two liquids are combine into a flask, then the flask feels hot when we touch it because the system released heat into the surrounding.

In general, exothermic process releases heat and rise the temperature of surrounding.

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Answer:

The molecular formula is C3H6

Explanation:

Step 1: Data given

Suppose the compound has a mass of 100 grams

The compound contains:

85.63 % C = 85.63 grams C

14.37 % H = 14.37 grams H

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles

Moles = grams / molar mass

Moles C = 85.63 grams / 12.01 g/mol

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Moles H = 14.37 grams / 1.01 g/mol

Moles H = 14.2 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 7.130 moles / 7.130 moles = 1

H = 14.2 moles / 7.130 moles = 2

The empirical formula is CH2

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Explanation:

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The chemical species which accept proton are called Brønsted-Lowry base.

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(b)

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Cl^- is Bronsted lowry base and HCl is its conjugate acid.

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(e)

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4 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

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4 years ago