Answer: -
1) 8.33 minutes
2) 118.39 in/ s
180.43 m/min
10.83 km/ hr
Explanation: -
Speed of light = 3 x 10⁸ m/s
Distance of the earth from the sun= 93 million miles
We know 1 million = 1,000,000
Also 1 mile = 1609 m
Distance of the earth from the sun= 93 million miles
= 93,000,000 miles.
= 1.5 x 
m
Time taken = 
= ![\frac{1.5 x [tex] 10^{11}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1.5%20x%20%5Btex%5D%2010%5E%7B11%7D%20%20)
m}{3 x 10⁸ m/s} [/tex]
= 500 s
= 500/ 60
= 8.33 minutes
2) Distance = 1 mile = 63360 inches
Time taken = 8.92 min
= 8.92 x 60
= 535.2 s
Speed = 
= 
= 118.39 in/ s
Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 
m
Time taken = 8.92 min
Speed =
= ![\frac{63360 x 2.54 x [tex] 10^{-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B63360%20x%202.54%20x%20%5Btex%5D%2010%5E%7B-2%7D%20%20)
m}{8.92 min} [/tex]
= 180.43 m/ min
1 m = 10⁻³ Km
1 min = 1/60 hour
1 m /min = 10⁻³ km/ 
= 60/1000
=0.06 km/hr
180.43 m / min = 180 x 0.06 km / hr
= 10.93 km / hr
The given reaction:
<span>ch3ch2cooh (aq) ↔ ch3ch2coo- (aq) + h+ (aq)
is called a reversible reaction.
This means that, the reaction does not reach an end point.
In this type of reactions, reactants react together to form products, while products combine together to form reactants.
So, the reaction proceeds in both direction forming both reactants and products.</span>
Salts that are from strong bases and strong acids do not hydrolyze. Salts that are from strong bases and weak acids do hydrolyze, which gives it a pH greater than 7. Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7.
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
A)The spring scale has a high level of precision and a low level of accuracy.
Explanation:
Hope it works for u guys
