The average number of televisions per family is numbered in the tens.
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Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is .
Hence, this is the required solution.
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>