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Aleksandr-060686 [28]

3 years ago

11

Which segement of the heating curve shown above represents an increase in the potential energy, but no change in the kinetic ene

rgy?
a. RS
b. TU
c. QR
d. PQ

2 answers:

Oliga [24]3 years ago

6 0

The answer is RS. Hope this helps!

Lisa [10]3 years ago

5 0

RS because it is moving but not a lot

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The oxidation of the sugar glucose, C6H12O6, is described by the following equation. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Ivahew [28]

Answer:

(a) 282 kJ

(b) 67.4 Calories

Explanation:

(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.

We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):

(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ

(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.

(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories

8 0

2 years ago

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A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2I

Daniel [21]

Answer: Thus the value of $K_{eq}$ is 110.25

Explanation:

Initial moles of $I_2$ = 0.500 mole

Initial moles of $Br_2$ = 0.500 mole

Volume of container = 1 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

equilibrium concentration of $IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M$ [/tex]

The given balanced equilibrium reaction is,

$I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)$

Initial conc. 0.500 M 0.500 M 0 M

At eqm. conc. (0.500-x) M (0.500-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}$

$K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}$

we are given : 2x = 0.84 M

x= 0.42

Now put all the given values in this expression, we get :

$K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}$

$K_c=110.25$

Thus the value of the equilibrium constant is 110.25

8 0

3 years ago