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4 years ago

In the measurement 543.1267 inches which digit is estimated

1 answer:

8 0

The total number of digits that are counted while performing a measurement, are called significant figures or significant numbers. In a measured number, all the digits are certain except the last digit. The last digit which is uncertain in a measured number is called estimated digit. In the measurement 543.1267 inches, the last digit 7 is estimated digit. While taking into account the total number of significant figures, we count all the certain digits plus the estimated digit. Therefore, the number 543.1267 has 7 significant figures.

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3 years ago

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Answer:

A, F, H

Explanation:

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The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed

Andru [333]

The question is incomplete, here is the complete question:

The decay constant for 14-C is $0.00012yr^{-1}$ In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.

Answer: The formula for the age of the charcoal is $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

Explanation:

Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 20) = 80 grams

Putting values in above equation, we get:

$1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

Hence, the formula for the age of the charcoal is $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

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3 years ago

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