Which best describes the changes occurring to the earth's surface over time?

Explain the components of Petroleum and their uses.

RideAnS [48]

<h2>answer:</h2>

Various Constituents of Petroleum and their Uses

Constituents of Petroleum

Kerosene

uses➪ Fuel for stoves, in lamps and as aviation fuel for jet aircrafts

Constituents of Petroleum

LPG

uses➪ Fuel for home and industry

Constituents of Petroleum

Lubricating oil

uses➪Used for Lubrication in machines and engines

Constituents of Petroleum

Paraffin wax

uses➪ Ointments, candles, vaseline etc.

8 0

3 years ago

Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

$HONH_2+HCl\rightarrow HONH_3^+Cl^-$

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

$V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL$

Now, we can see that the moles of acid, base and acidic salt are all:

$0.0220L*0.200mol/L=0.0044mol$

And therefore the concentration of the salt at the equivalence point is:

$[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M$

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

$HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2$

Whereas the equilibrium expression is:

$Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}$

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}} \\\\Ka=1.10x10^{-6}$

So we can write:

$1.10x10^{-6}=\frac{x^2}{0.0858-x}$

And neglect the x on bottom to obtain:

$1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M$

And since x=[H3O+] we obtain the following pH:

$pH=-log(3.07x10^{-4})\\\\pH=3.513$

Regards!

4 0

3 years ago

Lewis structures for the perchlorate ion (ClO4−) can be drawn with all single bonds or with one, two, or three double bonds. Dra

Vanyuwa [196]

Answer:

The most important resonance structure is 4 (attached picture). Its bon order is $\frac{7}{4}$ or $1\frac{3}{4}$ .

Explanation:

A picture with 4 forms of the perchlorate structure is attached. The first structure has simple bonds. The second structure contains a double bond, the third structure has two double bonds and the fourth structure has three double bonds.

Formal charge = group number of the periodic table - number of bonds (number of bonding electrons / 2) - number of non-shared electrons (lone pairs)

The formal charges in the first structure is +3 in chlorine and -1 in oxygen.

The formal charges in the second structure is +2 in chlorine, -1 in oxygen and 0 in the double bond oxygen.

The formal charges in the third structure is +1 in chlorine, -1 in the single bond oxygens and 0 in the double bond oxygens.

The formal charges in the fourth structure is 0 in chlorine, -1 in the single bond oxygen and 0 in the double bond oxygens.

The most important resonance structure is given by: