How many valence electrons does antimony (51Sb) have

How many valence electrons does antimony (51Sb) have?

a.

7

b.

2

c.

3

d.

5

e.

4

1.b

The correct answer for the addition of 7.5 g + 2.26 g + 1.311 g + 2 g is which of the following to the correct number of significant figures?

a.

13.071

b.

13

c.

13.1

d.

13.0

1.c

The age of an ancient tree trunk is estimated using radiocarbon dating. If the trunk has a C-14 decay rate that is 6.250% of what it is in living plants, how old is the trunk? The half-life of C-14 is 5,730 years

The Correct Answer and Explanation is :

1. Antimony (Sb) and its valence electrons:

Antimony (Sb) is in Group 15 (also called Group V) of the periodic table, which means it has 5 valence electrons. This is the key factor that determines its ability to form bonds and its chemical behavior.

Correct answer: d. 5

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2. Addition with correct significant figures:

When adding measurements, you need to consider the least number of decimal places. Here’s the breakdown:

• 7.5 g has 1 decimal place.
• 2.26 g has 2 decimal places.
• 1.311 g has 3 decimal places.
• 2 g has no decimal places.

For addition, the result should have the least number of decimal places, which in this case is 1 decimal place (from 7.5 g). So, rounding the sum to 1 decimal place, the correct result is:

13.1

Correct answer: c. 13.1

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3. Radiocarbon dating:

Radiocarbon dating helps determine the age of a sample by comparing the amount of C-14 remaining in the sample to the original amount in living organisms. C-14 decays over time, and its half-life (the time it takes for half of the C-14 to decay) is 5,730 years.

In this case, the trunk has only 6.250% of the original C-14, meaning 93.75% has decayed. To calculate the age, we need to figure out how many half-lives have passed for the C-14 to decay from 100% to 6.250%.

Using the formula for half-life:

[
\text{Fraction remaining} = \left( \frac{1}{2} \right)^n
]

Where n is the number of half-lives, and the fraction remaining is 0.06250 (since 6.250% is 0.06250 in decimal form). We solve for n:

[
0.06250 = \left( \frac{1}{2} \right)^n
]

Taking the natural logarithm of both sides:

[
\ln(0.06250) = n \ln\left(\frac{1}{2}\right)
]

[
n = \frac{\ln(0.06250)}{\ln(0.5)} \approx \frac{-2.7726}{-0.6931} \approx 4
]

So, 4 half-lives have passed.

Now, we multiply the number of half-lives by the length of each half-life to find the age:

[
\text{Age} = 4 \times 5730 = 22,920 \text{ years}
]

Thus, the tree trunk is estimated to be about 22,920 years old.