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
Answer:
The molecular formula of the compound :
Explanation:
The empirical formula of the compound =
The molecular formula of the compound =
The equation used to calculate the valency is :
We are given:
Mass of molecular formula = 86 g/mol
Mass of empirical formula = 43 g/mol
Putting values in above equation, we get:
The molecular formula of the compound :
Al2(SO4)3 + 3Ca(OH)2 -> 2Al(OH)3 + 3Ca(SO)4
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
