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ArbitrLikvidat [17]
3 years ago
11

A single penny is 1.52 mm thick. The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years.

If it were possible to stack one mole of pennies, how many times would the stack go between the earth and Alpha Centauri? Use the unit factoring method to determine the answer and show your work. You will need to find or look up the appropriate conversion factors to solve the problem. Your answer should be in scientific notation and have the correct number of significant figures in order to get full credit. (Please note that the text editing functions/buttons below for this essay question allows you to show exponents by using the button show as "x2"in the controls. To use it type the number followed by the exponent such as 104, highlight the 4 and hit the x2 button and you will end up with 104 as the result)
Chemistry
1 answer:
crimeas [40]3 years ago
8 0

Answer:

2.29 × 10⁴ times

Explanation:

A single penny is 1.52 mm thick. The distance covered by 1 mole of pennies (6.02 × 10²³ pennies) is:

6.02 × 10²³ p × (1.52 mm/1 p) = 9.15 × 10²³ mm = 9.15 × 10²³ × 10⁻³ m = 9.15 × 10²⁰ m

The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. Considering 1 ly = 9.46 × 10¹⁵ m, this distance in meters is:

4.22 ly × (9.46 × 10¹⁵ m/1 ly) = 3.99 × 10¹⁶ m

The times that the stack would go between the earth and Alpha Centauri are:

9.15 × 10²⁰ m / 3.99 × 10¹⁶ m = 2.29 × 10⁴

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a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g
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Answer:

(A) \Delta H^{\circ }_{r}= -144 kJ

(B) \Delta H^{\circ }_{r}= - 2552kJ

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)

1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)  

Now, multiplying equation (a) with 2:

⇒ N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)

Then equation b is reversed and multiplied with 2:

2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)

Now by adding the equation (a) and equation (b), we get:

⇒  2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)

⇒  2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}

= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)    

B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)

2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)

H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)

H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)  

4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)

6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)

6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }

= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ

4 0
3 years ago
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