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3 years ago

What particle is emitted when lead-214, 214 82 pb undergoes natural radioactive decay to bismuth-214, 214 83 bi?

1 answer:

const2013 [10]3 years ago

6 0

A beta particle.β
 Looking at the particles before and after for the decay, you'll see that there is a total of 214 neutrons and protons both before and after the radioactive decay. But you'll also see that before the decay, there were 82 protons and after the decay, there are 83 protons. So a neutron was converted to a proton which means that a negative charge had to be emitted, so it's a beta decay where a neutron emits a beta particle (an electron). The beta particle can be represented as a greek beta β followed by a minus sign, giving β-, or as a "e" followed by a minus sign, so e-

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3 years ago

Photosynthesis allows plants to turn light energy into chemical energy by forming glucose from carbon dioxide and water: 6CO2(g)

dezoksy [38]

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305 g of CO₂

3.77 × 10⁵ kJ

Explanation:

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A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato?

Let's consider the following relations:

The potato is 100%-73%=27% glucose by mass.
1 lb = 453.59 g.
6 moles of CO₂ produce 1 mole of glucose.
The molar mass of glucose is 180.156 g/mol.
The molar mass of carbon dioxide is 44.01 g/mol.

Then, for a 1.70 lb potato:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{6molCO_{2}}{1molGlucose} .\frac{44.01gCO_{2}}{1molCO_{2}} =305gCO_{2}$

How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%?

According to the enthalpy of the reaction, 2802.8 kJ are required to produce 1 mole of glucose. Then, for a 1.70 lb potato:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{2802.8kJ}{1molGlucose} .\frac{1}{0.86\% } =3.77\times 10^{5} kJ$

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3 years ago