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borishaifa [10]
2 years ago
10

After 17.1 thousand years, what percentage of the original carbon-14 would be left in an organism’s remains? After 17.1 thousand

years, what percentage of the original carbon-14 would be left in an organism’s remains? 25% 17.1% 12.5% 6.25%
Chemistry
1 answer:
hram777 [196]2 years ago
6 0

Answer:

12.5%

Explanation:

Initial percentage of carbon 14 = 100%

Final percentage = ?

Time passed = 17.1 * 1000 years = 17100 years

Half life of carbon 14 = 5,730 years.

So how many Half lives are in 17100 years?

Number of Half lives = Time passed /  Half life = 3.18 ≈ 3

First Half life;

100% --> 50%

Second Half life;

50% --> 25%

Third Half life ;

25% --> 12.5%

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2 years ago
what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg
zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

 ⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F =  3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0
3 years ago
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