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3 years ago

Calculate the pH of a solution with [OH−]=1.3×10−2M. (Hint: Begin by using Kw to find [H3O+].)

Chemistry

1 answer:

Nookie1986 [14]3 years ago

3 0

Answer:

12.11.

Explanation:

∵ pOH = - log[OH⁻]

∵ pH + pOH = 14.

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How many neutrons does 3919K have?

Igoryamba

19 protons and 19 electrons but neutrons are vary

5 0

3 years ago

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed

soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

KClO₃: 2 moles
KCl: 2 moles
O₂: 3 moles

The molar mass of the compounds is:

KClO₃: 122.45 g/mole
KCl: 74.45 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
KCl: 2 moles ×74.45 g/mole= 148.9 grams
O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that the gas occupies.
T is the temperature of the gas.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
V= 165 mL= 0.165 L (being 1000 mL= 1 L)
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$ × 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$ × 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0

2 years ago

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

8 0

3 years ago

Why are valence electrons important?

Usimov [2.4K]

They are the outer layer of the electron layers.

5 0

3 years ago

Read 2 more answers

The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-catalyzed aldol reaction

svp [43]

Answer:

The reactive nucleophile is Ketone.

Explanation:

In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.

The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.

Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles

4 0

3 years ago