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suter [353]
3 years ago
10

1) You are asked to make 10 miles of iron (Fe) from iron oxide (Fe2O3) and excess carbon monoxide (CO). Fe2O3(s) + 3CO(g)—> 2

Fe(l) + 3O2(g). How many moles of iron oxide must you use.
A) 3 moles
B) 2 moles
C) 10 moles
D) 5 moles

2) 2C4H10 + 13O2–> 8CO2 + 10H2O. If I want to produce 50g of H2O using the above combustion reaction, how many mol of C4H10 should I use?

A) 0.55 mol C4H10
B) 10 mol C4H10
C) 2.77 mol C4H10
D) 3.79 mol C4H10

Chemistry
1 answer:
Inessa05 [86]3 years ago
8 0

For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.

1) D 5 mols

2) A 0.55 mols

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3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

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The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

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