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Aleks04 [339]
4 years ago
6

Why are the trends and exceptions to the trends in ionization energy observed? Check all that apply. -Ionization energy tends to

increase down a group because the electrons get farther away from the nucleus. -Ionization energy tends to increase across a period because the nuclear charge increases. -Ionization energy tends to increase across a period because electrons are added to the same main energy level. -The ionization energies of the elements in Group 16 tend to be slightly smaller than the elements in Group 15 because the fourth electron is added to an unfilled p orbital. -The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.
Chemistry
1 answer:
Annette [7]4 years ago
3 0

Answer:

Trends across a Period ... properties to structure, the chemical changes, the trends and patterns in the Periodic ... Strong emphasis will be placed on chemical energy changes to finally ... charged particles (ions) or groups of atoms (molecules). ... As the magnetic field is gradually increased, the separated ions are detected ...

Explanation:

Trends across a Period ... properties to structure, the chemical changes, the trends and patterns in the Periodic ... Strong emphasis will be placed on chemical energy changes to finally ... charged particles (ions) or groups of atoms (molecules). ... As the magnetic field is gradually increased, the separated ions are detected ...

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Considering the chart shown, at what temperature does the substance boil? 200ºC 1,200ºC 2,200ºC 2,400ºC
kodGreya [7K]

Answer:

The answer is 2,200ºC

Explanation:
I took the assignment for Edge, I don't think I can send the image because it might pick up on that and get reported, sorry!

The only other rationale that I have is that it's boiling because the graph shows that it's at a constant temperature/rate at 2,200ºC for quite a while. Typically when something boils, it stays at that constant rate of boiling, unless you turn the temperature up or it's finally able to peak..?

8 0
3 years ago
Which statement is false?
emmasim [6.3K]

Answer:

  • <u><em>The first statement is false:  a.At equilibrium, equal amounts of products and reactants are present. ΔG° is a function of Keq.</em></u>

Explanation:

When one part of a statement is false, the whole statement is false.

At <em>equilibrium,</em> the amounts of products and reactants does not have to be equal.

At equlibrium the rates of the forward reaction and the reverse reaction must be equal.

An equilibrium reaction may be represented by:

  • A + B ⇄ C + D

That represents two reactions:

  • Direct reaction: A + B → C + D (A and B yield C and D)

  • Reverse reaction: A + B ← C + D (C and D yield A and B: note that the arrow goes from right to left)

So, it is when the direct and the forward rates are equal that there is not net change in the amounts of all the species and so the reaction is is equilibrium).

As per the other statement, both parts are true:

  • When reactants become products, they do so through an intermediate transitrion state: when the reactants approach each other and collide with enough energy and appropiate position, the bonds start to break and the bonds of the products start to form. This is the transition state.

  • Most biocatalysts are proteins: enzymes are simply proteins, with specific structures, that may accelerate or even deceralate biochemical reactions.
5 0
3 years ago
Water supplies are often treated with chlorine as one of the processing steps in treating wastewater. Estimate the liquid diffus
jekas [21]

Answer:

⇒D_AB= 1.21×10^(-9)

Explanation:

Wike chang  equation is given as:

D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}

Where

D_AB= diffusivity of chlorine in water

Φ= 2.26 for water as solvent

ν= 0.0484 for chlorine as solute

M_B = Molecular weight of water

τ= temperature=289 K

μ= viscosity = 1.1×10^{-3}

Now putting these values in the above equation we get

D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}

⇒D_AB= 1.21×10^(-9)

7 0
3 years ago
Read 2 more answers
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
How many full orbitals are in phosphorus
andrew-mc [135]

Answer:

three half-filled orbitals

8 0
3 years ago
Read 2 more answers
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