Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton

Chemistry

1 answer:

MrRa [10]3 years ago

4 0

Answer: Option (c) is the correct answer.

Explanation:

A positron is denoted as $^{0}_{+1}e$ , a neutron is denoted as $^{1}_{0}n$ , an electron is denoted as $^{0}_{-1}e$ , and a proton is denoted as $^{1}_{1}p$ .

Therefore, when $^{219}Rn$ will decay into $^{218}Rn$ this means there occurs decrease in its mass number. Hence, it means a neutron will be releasing during this reaction.

The reaction will be as follows.

$^{219}Rn \rightarrow ^{218}Rn + ^{1}_{0}n$

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3 years ago

Consider the process used to produce iron metal from its ore.

Dmitry_Shevchenko [17]

Answer:

223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.

Explanation:

The balanced reaction is:

Fe₂O₃ (s) + 3 CO(g) → 2 Fe(s) + 3 CO₂ (g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

Fe₂O₃: 1 mole
CO: 3 moles
Fe: 2 moles
CO₂: 3 moles

Being:

Fe: 55.85 g/mole
O: 16 g/mole
C: 12 g/mole

the molar mass of the compounds participating in the reaction is:

Fe₂O₃: 2*55.85 g/mole + 3*16 g/mole= 159.7 g/mole
CO: 12 g/mole + 16 g/mole= 28 g/mole
Fe: 55.85 g/mole
CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole

Then, by stoichiometry of the reaction, the following quantities participate in the reaction:

Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 g
CO: 3 moles* 28 g/mole= 84 g
Fe: 2 moles* 55.85 g/mole= 111.7 g
CO₂: 3 moles* 44 g/mole= 132 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

So, first of all, you can apply the following rule of three: if by reaction stoichiometry 1 mole of Fe₂O₃ reacts with 3 moles of CO, then 2.5 moles of Fe₂O₃ react with how many moles of CO?

$moles of CO=\frac{2.5 moles of Fe_{2} O_{3}*3 moles of CO }{1 mole of Fe_{2} O_{3}}$

moles of CO= 7.5

But 7.5 moles of CO are not available, 6.0 moles are available. Since you have less moles than you need to react with 2.5 moles of Fe₂O₃, CO will be the limiting reagent.

Now you can apply the following rule of three: if by reaction stoichiometry 3 moles of CO produce with 111.7 grams of Fe, then 6 moles of CO will produce how much mass of Fe?

$mass of Fe=\frac{6 moles of CO*111.7 grams of Fe}{3 moles of CO}$