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kozerog [31]
3 years ago
8

A solution with pH of 9 has [OH-] concentration of

Chemistry
1 answer:
Drupady [299]3 years ago
7 0
PH = -log([H+])
[H+] = 10^(-pH)

[H+] = 10^(-9)

[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.

[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5

The concentration of OH- ions is 1.0*10^-5 M.
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Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Co
lord [1]

Solution :

$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$

<u>Explanation 1 </u>:

Spontaneity of the reaction is based on two factors :

-- the tendency to acquire a state of minimum energy

-- the energy of a system to acquire a maximum randomness.

Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.

<u>Explanation 2 </u>:

A system containing the \text{"chemically mixed"} B halides has a \text{greater entropy} than a system of $BCl_3$ and BF_3.

It has the same number of \text{gas phase molecules}, but more distinguishable kinds of \text{molecules}, hence, more microstates and higher entropy.

8 0
3 years ago
What pressure (in kilopascals) is exerted by 4.20 moles of Xenon gas in a 15.0 L container at 280.0 K?
aniked [119]

Answer: the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa

Explanation:

The question requires us to calculate the pressure, in kPa, connsidering the following information:

<em>number of moles = n = 4.20mol</em>

<em>volume of gas = V = 15.0L</em>

<em>temperature of gas = T = 280.0 K</em>

We can use the equation of ideal gases to calculate the pressure of the gas, as shown by the rearranged equation below:

PV=nRT\rightarrow P=\frac{nRT}{V}

Since the volume was given in L and the question requires us to calculate the pressure in kPa, we can use R in units of L.Pa/K.mol:

<em>R = 8314.46 L.Pa/K.mol</em>

Applying the values given by the question to the rearranged equation above, we'll have:

\begin{gathered} P=\frac{nRT}{V} \\  \\ P=\frac{(4.20mol)\times(8314.46L.Pa/K.mol)\times(280.0K)}{(15.0L)}=652\times10^3Pa \end{gathered}

Therefore, the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa.

8 0
1 year ago
What are the list of catalysts that can be used for hydrogenation of nitrobenzene???​
Alex Ar [27]

Answer:

The multiring aromatic hydrocarbons in the coal liquid were hydrogenated to give saturated molecules that contained only one aromatic ring. Of the several organic bases investigated, potassium bis(trimethylsilyl)amide was found to be the most effective catalyst.

Explanation:

5 0
3 years ago
A 1.000 g sample of an unknown hydrate of cobalt chloride is gently dehydrated. The resulting mass is 0.546 g. The cobalt is iso
ira [324]

Answer:

Explanation:

From the 1 g sample you have:

0.546 grams of cobalt chloride

1-0.546=0.454 grams of water

Now:

1) The salt

Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):

n_{Co}=\frac{0.248 g}{58.9 g/mol}=4.21*10^{-3}mol

n_{Cl}=\frac{0.298 g}{35.45g/mol}=8.4*10^{-3}mol

Dividing:

\frac{n_{Cl}}{n_{Co}}=\frac{8.4*10^{-3}mol}{4.21*10^{-3}mol}=2

So the molecular formula will be:

CoCl_2

2) The water

The water's molecular weight is M=18 :

n_{w}=\frac{0.454g}{18g/mol}=0.025 mol

Bonding with the Co:

\frac{n_{w}}{n_{Co}}=\frac{0.025mol}{4.21*10^{-3}mol}=6

The complete formula of the hydrate:

CoCl_2*6 H_2O

8 0
4 years ago
I only need help in the bottom section? Please someone help me please?
kherson [118]
Energy in the nucleus of an atom of uranium: Nuclear energy
Energy of a moving object: Kinetic Energy 
Energy stored in chemical bonds: Potential Energy
Energy emitted from light bulb: Radiant Energy
Energy in a battery: Stored Energy
Energy in our food: Chemical Energy
Energy emitted from a radio: Electromagnetic Energy
Energy affected by mass and speed: Kinetic Energy
Energy affected by position and condition: Potential Energy
Energy from our star that some homes use for electricity: Radiant Energy
I hope this was found helpful!
I was happy to assist you with your Homework :)
4 0
4 years ago
Read 2 more answers
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