Answer:
the normality of the given solution is 0.0755 N
Explanation:
The computation of the normality of the given solution is shown below:
Here we have to realize the two sodiums ions per carbonate ion i.e.
N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)
= 0.1886eq ÷ 0.2500L
= 0.0755 N
Hence, the normality of the given solution is 0.0755 N
The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
brainly.com/question/26374513
14. C
15. B
16. D
17. D
18. I am not sure sorry.<span />
Kr= krypton
K= potassium
C=carbon
En= neon
Si= silicon
Au= gold
Ni= nitrogen
Br= berillium
Mg=magnesium
Mn= mangenese
Al: aluminum
