Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
Answer:
XY₂Z₄
2.35 mol Z
Explanation:
A sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z. We can find the simplest formula (empirical formula) by <em>dividing all the numbers of moles by the smallest one</em>.
X: 0.221/0.221 = 1
Y: 0.442/0.221 = 2
Z: 0.884/0.221 = 4
The simplest formula is XY₂Z₄.
The molar ratio of X to Z is 1:4. The moles of Z in a sample that contained 0.588 moles of X is:
0.588 mol X × (4 mol Z/1 mol X) = 2.35 mol Z
Answer:
Element Symbol Mass Percent
Cuprum Cu 66.464%
Sulfur S 33.537%
Explanation:
I got this out of my module, sorry if it's wrong but i am pretty sure 97% this is correct!
B and e
first we need to balance the NH3 hence first we do E and multiplying the coefficient by 2. that will leave us with N2+H2–>2NH3.
N2 and H2 is balanced and now all that is left to do is the balance H2 by 3 as there is 6H on RHS hence we need 6H on LHS
