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stellarik [79]
2 years ago
6

A 0.0884 M solution of a weak base has a pH of 11.79. What is the identity of the weak base?Weak Base KbEthylamine (CH3CH2NH2) 4

.7 x 10-4Hydrazine (N2H4) 1.7 x 10-6Hydroxylamine (NH2OH) 1.1 x 10-8Pyridine (C5H5N) 1.4 x 10-9Aniline (C6H5NH2) 4.2 x 10-10A) hydrazineB) pyridineC) anilineD) ethylamineE) hydroxylamine
Chemistry
1 answer:
iVinArrow [24]2 years ago
7 0

Answer:

The base must be ethylamine.

Explanation:

The pH of solution of a weak base gives us an idea about the Kb of the base.

pOH=14-pH

pOH=14-11.79=2.21

pOH=-log[OH^{-}]

[OH^{-}]=0.0062M

The relation between Kb and hydroxide ion concentration is:

Kb=\frac{[OH^{-}]^{2}}{[base]}

Kb=\frac{0.0062X0.0062}{0.0884}=4.34X10^{-4}

Thus the weak base must be ethylamine.

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3.80 moles of oxygen are used up in the reaction. How many moles of water are produced?5.How many grams of oxygen does it take t
r-ruslan [8.4K]

Answer:

2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)

2.5 g x g

Explanation:

x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2

(122.5 g KClO3) (2 mol KClO3) (1 mol O2)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

2.5 g x g

x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl

(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

x mol 10 mol

x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3

(3 mol O2)

7 0
2 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
Predict the precipitate produced by mixing an Al(NO3)3 solution with a NaOH solution. Write the net ionic equation for the react
weqwewe [10]
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)

The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
4 0
3 years ago
For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati
aleksandrvk [35]

Answer:

Br- Withdraws electrons inductively

       Donates electrons by resonance

CH2CH3 - Donates electrons by hyperconjugation

NHCH3- Withdraws electrons inductively

              Donates electrons by resonance

OCH3 -  Withdraws electrons inductively

              Donates electrons by resonance

+N(CH3)3 - Withdraws electrons inductively

                   

Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.

Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

-NHCH3  and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.

+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.

3 0
3 years ago
How do scientists take instruments for measurement up to about 100,000 feet of
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Answer: b
can’t be weather balloons cuz they would pop.
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4 0
2 years ago
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