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3 years ago

What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?

Chemistry

2 answers:

bekas [8.4K]3 years ago

6 0

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Over [174]3 years ago

5 0

Answer:

the answer is 8.02

Explanation:

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Read 2 more answers

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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3 years ago