Answer:
<u>1.4 M</u>
Explanation:
n(HBr)=3.115/81
so, 3.115/81=0.0385mol
according to the reaction, n(HBr)=n(KOH)=0.038 mol
C(KOH)=n/V=0.0385/0.02745
0.0385/0.02745 =1.4 M
in short the answer is 1.4 M (molarity)
Answer:
0.6941 mg
Explanation:
First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).
Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
5 is 2 I’m not sure about 4 though....
Answer: In the first paragraph, name a theme of Paul Laurence Dunbar's poem "Sympathy," and explain how it develops, citing specific examples
Explanation:
