All categories

2 years ago

Why is the bond between two single hydrogen atoms stable?

Chemistry

2 answers:

Svetlanka [38]2 years ago

6 0

Answer:

B. Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.

Explanation:

As hydrogen is the first element in the periodic table and has the atomic number it can mix with a variety of elements, including itself, when two atoms of Hydorgen bond the bond they create fills up the first level of energy which only uses 2 electrons in order to be completed, which makes the bond stable by filling up the energy level.

Len [333]2 years ago

5 0

B ia the correct one .

Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.

You might be interested in

How many moles are there in 7.24 grams of calcium carbonate? With work shown

balu736 [363]

Answer:

0.0723371390261859

8 0

2 years ago

Read 2 more answers

An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?

juin [17]

Answer:

$pKa=3.58$

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

$pH=-log([H^+])$

$[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}$

And the percent ionization is:

$\% \ ionization=\frac{[H^+]}{[HA]}*100\%$

We compute the concentration of the acid, HA:

$[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\$

$[HA]=6.03x10^{-4}$

Thus, the Ka is:

$Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}$

So the pKa is:

$pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58$

Regards.

5 0

2 years ago

Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.

mojhsa [17]

Answer:

mass = 100 g

T1 = 0°C

T2 = 100 °C

C = 1 cal/g°C

Q = mC(T2 -T1)

Q = 100(1)(100 - 0)

Q = 100(100)

Q = 10000 cal

Explanation:

6 0

2 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

Why this process is called a cycle

Greeley [361]

<span>because it is a pattern; airgo cycle</span>

5 0

3 years ago

Read 2 more answers