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Allisa [31]
3 years ago
11

Read, now please rail me jkjkjk.................Unless

Chemistry
2 answers:
Ludmilka [50]3 years ago
8 0

Answer:

no... can some 1 answer my last two questions they are the same so you get 20 pts !!!!!!!

Explanation:

algol133 years ago
7 0

I- I'm a minor

but ummm jfuibhiuhdcuihshuihih

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Rank the following elements by electron affinity, from most positive to most negative EA value. Rank from most positive to most
Svetlanka [38]

Answer:

Most positive = rubidium

Most negative = fluorine

Explanation:

Electron affinity of an element is the energy released when an electron is attached to a neural atom to form an ion in its gaseous state.

X + e⁻ → X⁻

Electron affinity is similar to electronegativity which is the tendency at which an atom accepts an ion towards itself.

Electron affinity increases across the period and decreases down the group in the periodic table.

In the above option,

Fluorine has the highest electron affinity

Rubidium has the lowest electron affinity

Tellurium and then finally Phosphorus

Helium in this case would have the lowest electron affinity because it has filled orbital and does not require any electron to attain stability. Technically, Helium has the lowest or is expected to have the lowest electron affinity which is close to zero according to quantum mechanics.

Most positive = rubidium

Most negative = fluorine.

You can check periodic table for their exact values

6 0
3 years ago
What is electrons configuration for arsenic ?
Anna35 [415]

Answer:

[Ar] 3d¹⁰ 4s² 4p³

Explanation:

so...... ur welcome

7 0
2 years ago
You prepare a 2nd solution (solution B) by pipetting 10.00 ml of your 1st solution (solution A) into a 50.00 ml volumetric flask
elena-14-01-66 [18.8K]

Answer:

Concentration solution A was 0.5225 M

Explanation:

10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B

According to laws of dilution-      C_{A}V_{A}=C_{B}V_{B}

where, C_{A} and C_{B} are concentration of solution A and B respectively

V_{A} and V_{B} are volumes of solution A and B respectively

Here C_{B} = 0.1045 M, V_{B} = 50.00 mL and V_{A} = 10.00 mL

Hence, C_{A}=\frac{C_{B}V_{B}}{V_{A}}=\frac{(0.1045M\times 50.00mL)}{10.00mL}=0.5225M

So, concentration solution A was 0.5225 M

8 0
3 years ago
Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.
Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

            HNO3   ↔    H+      +      NO3-

I            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0
3 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
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