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kolbaska11 [484]
3 years ago
6

A) For the following reaction, K > 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry a

cids or bases.
C9H7N + HNO2Doublearrow.GIFC9H7NH+ + NO2-

a) HNO2 1) stronger acid

b) NO2- 2) weaker acid

c) C9H7NH+ 3) stronger base

d) C9H7N 4) weaker base

B)
For the following reaction, K < 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry acids or bases.

C5H11NH+ + C6H5COO-Doublearrow.GIFC5H11N + C6H5COOH

a) C5H11NH+ 1) strongest acid

b) C6H5COO- 2) stongest base

c) C5H11N 3) weakest acid

d) C6H5COOH 4) weakest base
Chemistry
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

Explanation:

According to Bronsted-Lowry acids or base theory ,  the reagent capable of giving hydrogen ion or proton  will be acid and that which accepts hydrogen ion or proton  will be base .

C₉H₇N + HNO₂   ⇄    C₉H₇NH⁺ + NO₂⁻

If K > 1 , reaction is proceeding from left to right .

Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form

C₉H₇NH⁺ .  

Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .

B )

when K < 1 , reaction above proceeds from right to left . That means

C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .

Hence ,  NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .

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under certain water​ conditions, the free chlorine​ (hypochlorous acid,​ hocl) in a swimming pool decomposes according to the la
ivolga24 [154]

Answer:

1.7 ppm

Explanation:

Original amount N' = 2.6 ppm

time to testing t = 24 hr

final amount N = 2.1 ppm

Using exponential inhibited decay, we have

N = N'e^(-kt)

Where

N is the new reading

N' is the original reading

t is the decay time

k is the decay constant

Substituting, we have

2.1 = 2.6 x e^(-k x 24)

2.1 = 2.6 x e^(-24k)

0.808 = e^(-24k)

We take the natural log of both sides of the equation

Ln 0.808 = Ln (e^(-24k))

-0.213 = - 24k

K = 0.213/24 = 0.00886

After 48 hrs, the reading of free chlorine will be

N = 2.6 x e^(-0.00886 x 48)

N = 2.6 x e^(-0.425)

N = 2.6 x 0.654

N = 1.7 ppm

5 0
3 years ago
Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
The nucleus is like the____of the cell
Tpy6a [65]

Answer:

Brain

Explanation:

It tells the other organelles how to do things

5 0
3 years ago
Read 2 more answers
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
Most elements are ____________. The _____________ elements on the left of the stair-step line on the ______________ of Elements
Vlada [557]

Answer:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster  , malleable , ductile, opposite , shiny, moderately

Explanation:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster  , malleable , ductile, opposite , shiny, moderately Is the answer.

(Hope this helps can I pls have brainlist (crown)☺️)

4 0
2 years ago
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