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3 years ago

What's the highest electronegativity

Chemistry

2 answers:

Bingel [31]3 years ago

5 0

Thus, fluorine is the most electronegative element, while francium is one of the least electronegative. (Helium, neon, and argon are not listed in the Paulingel ectronegativity scale, although in the Allred-Rochow scale, helium has the highest electronogativity.)

TiliK225 [7]3 years ago

3 0

Your answer is H) florine

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Solid magnesium reacts with aqueous Iron(III) chloride to produce aqueous magnesium chloride and solid iron.

luda_lava [24]

Answer:

9÷2Mg2+2FeCl3 =3Mg3Cl2 +2Fe

3 0

4 years ago

You make 1 Liter of an aqueous solution containing 9.20 ml of 57.8 mM acetic acid and 56.2 mg of sodium acetate (MW = 82.0 g/mol

Whitepunk [10]

Answer:

a) 5,3176x10⁻⁴ moles

b) 6,85x10⁻⁴ moles

c) The appropriate formula to calculate is Henderson-Hasselbalch.

d) pH = 4,86. Acidic solution but slighty

Explanation:

a) moles of acetic acid:

9,20x10⁻³L × 57,8x10⁻³M = 5,3176x10⁻⁴ moles

b) moles of sodium acetate:

56,2x10⁻³g ÷ 82,0 g/mole = 6,85x10⁻⁴ moles

c) The appropriate formula to calculate is Henderson-Hasselbalch:

pH= pka + log₁₀ $\frac{[A^-]}{[HA]}$

d) pH= 4,75 + log₁₀ $\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}$

pH = 4,86

3 < pH < 7→ Acidic solution but slighty

I hope it helps!

3 0

4 years ago

This solution contains equal concentrations of both hf(aq) and naf(aq). write the proton transfer equilibrium (net ionic) reacti

Viktor [21]

Net ionic reaction for the solution containing equal concentrations of both HF(aq) and NaF(aq) is as follow:
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)

3 0

3 years ago

Read 2 more answers

Consider the following data showing the initial rate of a reaction (A→products) at several different concentrations of A.

OlgaM077 [116]

Answer:

(1) order = 2

(2) R = K [A]²

Explanation:

Given the reaction:

A--------->Product

The rate constant relation for the reaction is given as:

R(i) = K [A]............(*)

Where R(I) is rate constant at different concentration of A.

Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results

0.011 = K [0.15] ⁿ.........(1)

0.044 = K [0.30]ⁿ .......(2)

0.177 = K [0.60]ⁿ .........(3)

Dividing (2) by (1) and (3) by (1)

Gives:

0.044/0.011 = [0.3/0.15]ⁿ

4 = 2ⁿ; 2² = 2ⁿ; n = 2

Similarly

0.177/0.011 = [0.60/0.15]ⁿ

16.09 = 4ⁿ

16.09 = 16 (approximately)

4² = 4ⁿ ; n = 2

Hence the order of the reaction is 2.

The rate law is R = K [A]²

4 0

3 years ago

Select all that apply. Which of the following bases are strong enough to deprotonate CH3CH2CH2Câ¡CH (pKa = 25), so that equilibr

Trava [24]

Answer:

b, c, and d.

Explanation:

In an acid-base reaction, the equilibrium favors the weaker acid, thus, to the reaction favors it, the product must be weaker than the acid given. As higher is the value of pKa, as weak is the acid. Thus, the base must form a conjugate acid with pKa > 25.

The values of pKa can be found in equilibrium tables. The salts will dissociate to form the acids, which will gain one proton (H+)

a. CH3ONa -> Conjugate acid CH3OH, pKa = 15.5

b. NaNH2 -> Conjugate acid NH3, pKa = 38

c. NaH -> Conjugate acid H2, pKa = 35

d. CH3CH2Li -> Conjugate acid CH3CH3, pKa = 50

e. H2O -> Conjugate acid H3O+, pKa = -1.7

f. NH3 -> Conjugate acid NH4+, pKa = 9.2

Thus, the ones with pKa > 25 are NH3, H2, and CH3CH3, thus the bases are NaNH2, NaH, and CH3CH2Li.

5 0

4 years ago