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3 years ago

6

The elements found in Group 2A, from top to bottom, include beryllium, magnesium, calcium, strontium, and barium. Which of these

elements has the highest electronegativity value?

1 answer:

pashok25 [27]3 years ago

6 0

Answer is beryllium.

The electronegativity trend is that it increases from left to right in a period and from bottom to top in a group. This is why fluorine has the highest electronegativity out of all the elements. So since beryllium is at the very top of the group, it has the highest electronegativity

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Uranium has atomic number 92. Its most common isotope is 238U, but the form used in nuclear bombs and nuclear power plants is 23

Yakvenalex [24]

Explanation:

Atomic Number = Number of protons

Mass Number = Number of protons + Number of neutrons

Isotopes are simply atoms of an element with the same number of protons and different number of neutrons.

First Isotope -- 238U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons = 238 - 92 = 146

Second Isotope -- 235U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons = 235 - 92 = 143

3 0

3 years ago

If one adds 0.1 mol of the weak acid hf (pk_a = 3.2) to a solution with a ph = 2, which species would be most abundant and how m

zavuch27 [327]

$F^{-}$ is most abundant and 6310 times more than HF.

<h3>What is a strong and weak acid?</h3>

When an acid is dissolved in water, all of its molecules disintegrate, making the acid powerful.

When an acid is dissolved in water, only a small number of its molecules disintegrate, making the acid weak. Strong acids have a lower pH than weak acids.

The powerful acids include perchloric acid, chloric acid, nitric acid, sulfuric acid, hydrobromic acid, and hydroiodic acid.

Given:

Pka=3..2

pH=7

Let the volume be 1 liter

[HF]=01 M

$pH=pka+log \frac{F^{-}}{HF} \\\\7=3.2+log\frac{F^{-}}{HF} \\3.8=log\frac{F^{-}}{HF}\\ \frac{F^{-}}{0.1}=10^{3.8} \\F^{-}=630.95 M$

Now,

$\frac{F^{-}}{HF}=\frac{630.95}{0.1}\\ =6309.57$

F-:HF= 6309.57:1

Therefore, the most abundant is $F^{-}$ and has 6310 times more than HF is $F^{-}$ .

To know more about strong and weak acids, visit: brainly.com/question/12811944

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4 0

1 year ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

$Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)$

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

3 0

3 years ago

Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and

Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

The given data is as follows.

mass = 500 g, $C_{p}$ = 24.4 J/mol K

$T_{h}$ = 500 K, $T_{c}$ = 250 K

Mass number of copper = 63.54 g /mol

Number of moles = $\frac{mass}{/text{\molar mass}}$

= $\frac{500}{63.54}$

= 7.86 moles

Now, equating the entropy change for both the substances as follows.

$7.86 \times 24.4 \times [T_{f} - 250]$ = $7.86 \times 24.4 \times [500 -T_{f}]$

$T_{f} - 250 = 500 - T_{f}$

$2T_{f}$ = 750

So, $T_{f}$ = $375^{o}C$

For the metal block A, change in entropy is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

= $24.4 log [\frac{375}{500}]$

= -3.04 J/ K mol

For the block B, change in entropy is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

= $24.4 log [\frac{375}{250}]$

= 4.296 J/Kmol

And, total entropy change will be as follows.

= 4.296 + (-3.04)

= 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.

8 0

3 years ago