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castortr0y [4]
2 years ago
5

Traveling at the speed of light - how long would it take, in years, to reach the top of the pile of balls? Show all calculations

. The height of the pile is 1.10 x 10^14 km
Chemistry
1 answer:
marusya05 [52]2 years ago
7 0

The time taken for the object to reach to top of pile is 0.012 year.

<h3>Time of motion </h3>

The time taken for the object to reach to top of pile is calculated as follows;

time of motion = distance traveled/speed

time of motion = (1.1 x 10¹⁴ x 10³ m)/(3 x 10⁸ m/s)

where;

  • speed of light = 3 x 10⁸ m/s

time of motion = 3.67 x 10⁵ sec = 0.012 year

Thus, the time taken for the object to reach to top of pile is 0.012 year.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

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Opal is a hydrated form of silica. If a laboratory analysis of a sample of opal reveals it to contain29.2% Si, 33.3%O, and 37.5%
ZanzabumX [31]
Let us assume that there is a 100g sample of Opal. The masses of each element will be:
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33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
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8 0
4 years ago
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole
trapecia [35]

Answer : The enthalpy of combustion per mole of C_6H_{12}O_6 is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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

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

The equilibrium reaction follows:

C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)

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

\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol

Therefore, the enthalpy of combustion per mole of C_6H_{12}O_6 is -2815.8 kJ/mol

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Explanation:

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