Is there a relationship between an elements atomic number and its first ionization energy (IE)?

Answers

Answer 1
Answer: yes, well kind of, as you travel across the a period  the first ionization energy increases). as you travel down a column the ionization energy decreases.  

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in a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use mc018-1.jpg.

Answers

Ans: Volume of stock H2SO4 required = 6.94 ml

Given:

Concentration of stock H2SO4 solution M1 = 18.0 M

Concentration of the final H2SO4 solution needed M2 = 2.50 M

Final volume of H2SO4 needed, V2 = 50.0 ml

To determine:

Volume of stock needed, V1

Explanation:

Use the dilution relation:

M1V1 = M2V2\n\nV1 = (M2V2)/(M1) \n\nV1  = (2.50 M * 50.0 ml)/(18.0 M) = 6.94 ml

Hello!

In a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution?  

We have the following data:

M1 (initial molarity) = 2.50 M (or mol/L)

V1 (initial volume) = 50.0 mL → 0.05 L

M2 (final molarity) = 18.0 M (or mol/L)

V2 (final volume) = ? (in mL)

Let's use the formula of dilution and molarity, so we have:

M_(1) * V_(1) = M_(2) * V_(2)

2.50 * 0.05 = 18.0 * V_(2)

0.125 = 18.0\:V_2

18.0\:V_2 = 0.125

V_2 = (0.125)/(18.0)

V_2 \approx 0.00694\:L \to \boxed{\boxed{V_2 \approx 6.94\:mL}}\:\:\:\:\:\:\bf\green{\checkmark}

Answer:

The volume is approximately 6.94 mL

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\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

If the plum pudding model was correct, what would the result of rutherfords expirement be

Answers

Most of the positively charged particles should bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil.

For a particular reaction, ΔH = -144.6 kJ and ΔS = -301.2 J/K. Calculate ΔG for this reaction at 298 K.

Answers

The Gibbs free energy change for this reaction at 298 K is approximately -54,484.8 J.

Determining the Gibbs free energy change

Calculating the Gibbs free energy change (ΔG) for a reaction at a given temperature using the Gibbs-Helmholtz equation,

ΔG = ΔH - TΔS

ΔG = (-144,600 J) - (298 K)(-301.2 J/K)

ΔG = -144,600 J + 90,115.2 J

ΔG = -54,484.8 J

ΔG = -54,484.8 J

Converting to kilojoules:

ΔG = -54,484.8 J / 1,000

= -54.48 kJ

Learn more on Gibbs free energy change here brainly.com/question/13765848

#SPJ3

To calculate, I would need to know if the final answer is supposed to be in kJ or in J/K.

Either way, use the Gibb's Formula: 
ΔG= ΔH-TΔS

If the final answer is supposed to be in J/K then plug-in your values to the formula and simply solve.

However, if your final answer is supposed to be in kJ, your first step would be to convert 
ΔS from J/K into kJ. 

This is simply done by dividing the given ΔS by 1000 ----> (-301.2/1000).
(we divide by 1000 because by definition that's what a "kilo" is )

New 
ΔS value= -0.3012 kJ 

Now all you have to do is plug-in all the values into the formula. 
ΔG= ΔH-TΔS
ΔG= (-144.6 kJ) - 298K(-0.3012kJ)
= -54.84 kJ 

Since the ΔG is negative, this means that the reaction is spontaneous. 

Hope this helps! :)

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

Answers

We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_(1)  V_(1) = M_(2)  V_(2)

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_(2)  SO_(4) is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_(2)

We can then solve for V_(2):

V_(2)= ((18M)(0.05L))/(4.35M) and V_(2) =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


1 Iron can be extracted from iron(III) oxide using carbon, shown by the equation below:Fe²O³ + C -> Fe + CO
Balance the equation and state which substance is oxidised and which is reduced​

Answers

Explanation:

To balance the chemical equation for the extraction of iron from iron(III) oxide using carbon, you need to ensure that the same number of each type of atom appears on both sides of the equation. Here's the balanced equation:

2Fe₂O₃ + 3C -> 4Fe + 3CO

Now, let's determine which substance is oxidized and which is reduced:

1. Iron (Fe) undergoes a change in oxidation state from +3 in iron(III) oxide (Fe₂O₃) to 0 in elemental iron (Fe). It is reduced because it gains electrons, going from a higher positive oxidation state to zero.

2. Carbon (C) in carbon (C) undergoes a change in oxidation state from 0 to +2 in carbon monoxide (CO). Carbon is oxidized because it loses electrons and goes from an oxidation state of 0 to +2.

So, in this chemical reaction:

- Iron is reduced (it gains electrons).

- Carbon is oxidized (it loses electrons).

Is soap polar or no polar explain

Answers

soap is polar because it is not symmetrical in molecularly, this is because it must react with whats on your hand to fully clean it , dish soap is nonpolar/polar in order to combat grease which is also nonpolar, so nonpolar cleans nonpolar and polar cleans polar
Soap is a polar covalent bond because its electrons are not shared equally between the atoms.