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boyakko [2]
3 years ago
12

What is the classification of a solution of naoh with a ph of 8.3?

Chemistry
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

  • <u>Alkaline or basic solution </u>(alkaline and basic means the same)

Explanation:

According to the <em>pH</em>,  solutions may be classified as neutral, acidic, or alkaline (basic).

This table shows such classification:

pH               classification

  7                   neutral

> 7                   alkaline or basic

< 7                   acidic

Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).

Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.

  • pH = - log [H₃O⁺]

You can calculate the concentration of hydronium ions using antilogarithm properties:

pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501

NaOH solutions are alkaline solutions, bases, according to Arrhenius model,  because they contain OH⁻ ions and release them when ionize in water.

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Answer is: 0,453 <span> moles of oxygen will react </span><span>with 0.3020 moles of carbon(IV) oxide</span><span>.
</span>n(CO₂) = 0,302 mol.
From chemical reaction: n(CO₂) : n(O₂) = 4 : 6.
n(O₂) = 6 · 0,302 mol ÷ 4.
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Element R and Element Q have the same number of valence electrons.These elements both haves similar chemical behavior, but Eleme
vovangra [49]

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, element R and element Q having the same number of valence electrons means they belong to the same group in the periodic table which is the reason for there similar chemical behavior (as elements in the same group tend to have the same chemical properties).

Element R having fewer energy level (or electron shell) than element Q shows element R has fewer number of electrons than element Q and can be found earlier in the periodic table (or group in particular) when compared to element  Q

8 0
3 years ago
How much energy is required to convert 15.0 g of ice at −106 °C to water vapor at 125 °C? Specific heats are 2.09 J/g K for both
myrzilka [38]

Answer:

49.3 kJ of energy is required

Explanation:

An exercise of calorimetry at its best

First of all, convert the ice to water before melting.

Q = ice mass . C . ΔT

Q = 15 g . 2.09 J/g°C (0° - (-106°C)

15 g . 2.09 J/g°C . 106°C = 3323.1 J

Now we have to melt the ice, to change its state

Q = mass . latent heat of fusion

Q = 15 g . 0.335 kJ/g = 5.025 kJ .1000 = 5025 J

After that, we have liquid water at 0° and the ice has melted completely. We have to release energy to make a temperature change, to 100° (vaporization)

Q = 15g . 4.18 J/g°C (100°C - 0°C)

Q = 6270 J

Water has been vaporizated so we have to calculate, the state change.

Q = mass . latent heat of vap

Q = 15 g. 2.260 kJ/g

Q = 33.9 kJ (.1000) = 33900 J

Finally we have to increase temperature from 100°C to 125°C

Q = 15 g . 2.09 J/g°C . (125°C - 100°C)

Q = 783.75 J

To know how much energy is required to conver 15 g of ice, to water vapor at 125°C, just sum all the heat released.

3323.1 J + 5025 J + 6270 J + 33900 J + 783.75 J = 49301.85 joules.

Notice I have to convert kJ to J in two calcules to make the sum.

49301.85 joules / 1000 = 49.3 kJ

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