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Softa [21]
3 years ago
13

HELP!! b. At the equivalence point, all of the acid has been neutralized by the base. Why does the pH change so

Chemistry
1 answer:
CaHeK987 [17]3 years ago
3 0

At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.

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3 years ago
HELP ASAP !!!!!!Can someone tell me easy one line definitions of atomic number, mass number, and isotopes
rusak2 [61]

Answer:

Fundamental properties of atoms including atomic number and atomic mass. The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons.

Explanation:

Fundamental properties of atoms including atomic number and atomic mass. The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons.

5 0
2 years ago
Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What
coldgirl [10]

Answer:

\rm_{90}^{231}\text{Th}

Explanation:

The unbalanced nuclear equation is

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + X

Let's write X as a nuclear symbol.

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

 92 = 2 + Z , so  Z =   92 - 2 =  90

And your nuclear equation becomes

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} +\, _{90}^{231}\text{X}

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3 years ago
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3 years ago
The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is
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V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
5 0
3 years ago
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