Omg i lost everything ugh
To do it again
1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol
2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol
3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol
4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol
5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g
6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g
7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g
8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g
I cant believe i had to do this all over
Answer:
1. smaller. 2. smaller. 3. greater
Explanation:
1. H−O−H angle is 104.45 and H−C−H angle is 109.5
2. O−S−O angle is 119 and F−B−F angle is 120
3. The F−S−F bond angle in SF₆ is 90 and F−Br−F bond angle in BrF₅ is 84.8
Answer:
Constant volcanic eruptions Hope this helps!
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.
D is the answer. N, U, sat, J, mars, e, v, mer