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TEA [102]
3 years ago
11

Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.1 kW/m2 . How long does it take for 1.91

×109 J to arrive on an area of 1 m2 ?
Chemistry
1 answer:
Stels [109]3 years ago
3 0

Answer:

1.736 × 10⁶

Explanation:

Data provided in the question:

Intensity of the energy received from the sun = 1.1 kW/m² = 1.1 × 10³ W/m²

Energy  required to arrive = 1.91 × 10⁹ J

Area at the surface = 1 m²

Now,

we know that

Energy = Intensity × Area × Time

or

1.91 × 10⁹ = ( 1.1 × 10³ ) × 1 × Time

or

Time = [ 1.91 × 10⁹ ] ÷ [ 1.1 × 10³ ]

or

Time = 1.736 × 10⁶

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The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a
Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g

Carbon-14 is 1.6\times 10^{-10}\ \% of the carbon in the body. So,

\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g

Also,

14 g of Carbon-14 contains  6.023\times 10^{23} atoms of carbon-14

So,  

2.48\times 10^{-8}\ g of Carbon-14 contains  \frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8} atoms of carbon-14

Atoms of carbon-14 =  1.07\times 10^{15}

Given that:

Half life = 5730 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac{ln\ 2}{5730}\ years^{-1}

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = 3.154\times 10^7 s

So, The rate constant, k = \frac{0.00012}{3.154\times 10^7} s⁻¹ = 3.8\times 10^{-12}\ s^{-1}

Thus, decay events per second = K\times atoms decayed = 3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s = 4066 decay/s

4 0
3 years ago
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