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Jobisdone [24]
3 years ago
11

Technetium has a half-life of six hours. If you obtain 100g of a pure technetium sample, after 18 hours how much of the pure sub

stance will be remaining?
Chemistry
1 answer:
Butoxors [25]3 years ago
5 0

Answer: 12.5 grams.

Explanation:

You can find the amount of pure substance remaining by using the half-life formula:

Q_{1}=Q(\frac{1}{2})^{\frac{t}{t2} }

Where t is the amount of time, and t2 is the half life of the substance.

Substituting our values for the variables, we get this equation:

Q_{1} = 100(\frac{1}{2})^3\\\\Q_{1} = 100(\frac{1}{8})\\\\Q_{1} = 12.5

After 18 hours, there will be 12.5 grams of the substance remaining.

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4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

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3 years ago
Please help fast! I will give brainliest!
Kaylis [27]

Answer:(c)

Explanation:

3 0
3 years ago
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