M(NaCl) = Ar(Na) + Ar(Cl) · g/mol.
M(NaCl) = 22,99 + 35,45 · g/mol.
M(NaCl) = 58,44 g/mol.
V(NaCl) = 3,5 L.
m(NaCl) = 90 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 90 g ÷ 58,44 g/mol.
n(NaCl) = 1,54 mol.
c(NaCl) = n(NaCl) ÷ V(NaCl).
c(NaCl) = 1,54 mol ÷ 3,5 L.
c(NaCl) = 0,44 mol/l = 0,44 M.
Answer:
It is C. observed using the 5 senses.
Explanation:
you can observe a substance without changing its identity.
Answer:
3
Explanation:
At pH 3, the pentapeptide Ala-Glu-His-Val-Cys would contain two positively charged groups and one negatively charged group.
This is evident in that peptides have both carboxylic and amino groups.
Given that peptides have a certain amount of H⁺ and OH⁻ ions, it converts to zwitterions.
This conversion process is based on pKₐ and pKb of the acid and the base accordingly.
Therefore, they have two positively charged groups and one negatively charged grou
Answer:
Question Answer
The half-life of cobalt-60 is 5.26 years. How many half-lives have passed in 10.52 years? 2
12.5% of a radioactive sample are left. How many half-lives have passed? 3
After 3 half-lives, how much of a 400 gram sample of radioactive uranium remains? 50g
Explanation:
