Answer:
2.5 g mass of Ra-226 will remain after 4800 years.
Explanation:
Given data:
Mass of Ra-226 = 20 g
Sample remain after 4800 years = ?
Solution:
Half life of Ra-226 = 1600 years
Number of half lives passed = T elapsed / half life
Number of half lives passed = 4800 year / 1600 year
Number of half lives passed = 3
At time zero = 20 g
At first half life = 20 g/2 = 10 g
At 2nd half life = 10 g/2 = 5 g
At 3rd half life = 5 g/2 = 2.5 g
Thus, 2.5 g mass of Ra-226 will remain after 4800 years.
<span>0.929 g/ml
Density in this case is defined as mass per volume. Or in other words, mass divided by volume. So you simply divide 929 grams by 1000 ml. Giving
929 g / 1000 ml = 929/1000 g/ml = 0.929 g/ml</span>
Answer:
The answer is D- make sure you give good rate
To calculate<span> the average </span>atomic mass<span>, multiply the fraction by the </span>mass<span> number for each isotope, then add them together.</span>
The reaction equation:
2Li + O → Li₂O
Molar ratio of Li to Li₂O is:
2 : 1
So if 3.03 moles of Li are present:
2/1 = 3.03 / x
x = 1.515 moles of Li₂O will be produced.
