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andriy [413]
3 years ago
5

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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arsen [322]

Answer:

Only one seven carbon hydrocarbon was produced.

Explanation:

  • Alkynes are reduced completely in presence of Pt/H_{2} to produce alkanes.
  • Hydrogenation in presence of Pt is a simple nucleophilic addition process where H_{2} molecule adds onto an unsaturated bond.
  • Hept-1-yne, hept-2-yne and hept-3-yne are constitutionally isomeric to each other. Hence, after complete reduction, all three alkynes produced heptane as only product.
  • So, only one seven carbon hydrocarbon was produced.

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3 years ago
A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pr
harina [27]

Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

Explanation:

Given: T_{1} = 178 K,      P_{1} = 0.3 atm

T_{2} = 278 K,           P_{2} = ?

According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.

Formula used to calculate the pressure is as follows.

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\

Substitute the values into above formula is as follows.

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm

Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

3 0
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I might go with d it seems right to me hope this helps.

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The energy produced in the stars results from
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lora16 [44]
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