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lilavasa [31]
2 years ago
11

The indicator propyl red has ka = 3.3 × 10–6 . estimate the approximate ph range over which this indicator would change color.

Chemistry
1 answer:
ollegr [7]2 years ago
7 0
We let the chemical equation for the weak acid indicator propyl red HPr be
     HPr + H2O ↔ H3O+ + Pr-
Therefore, the acid dissociation constant Ka is
     Ka = [H3O+][Pr-] / [HPr]
The color of this indicator turns from red to yellow or the other way around at its turning point at which 
     [HPr] = [Pr-]
Substituting this to the equation for Ka, we now have
     Ka = [H3O+]
The pH of the solution at its turning point is 
     pH = -log[H3O+] = -log(Ka) = -log 3.3×10^-6 = 5.48
<span>On this account, the pH range is pH 5.0 to 6.0.</span>
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One hour of bicycle riding can require 500-900 kcal of energy, depending on the speed, the terrain, and the weight of the racer.
solniwko [45]

Answer:

145 hours

Explanation:

Since one hour of riding a bicycle takes up 505 kcal of energy. It is also stated that one gram of body fat is equal to 7.70 kcal. Also, it is given that 1 pound of body fat is equal to 454 g.

Hence;

1 Ib= 454 g

21 Ib= 21 × 454/1 = 9534 g

But

1g of body fat = 7.70kcal

9534 g of body fat = 9534 × 7.70 kcal/1 = 73411.8 kcal

If 505 kcal is lost in 1 hour

73411.8 kcal is lost in 73411.8 kcal × 1hour/505k cal = 145 hours

3 0
2 years ago
A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
GrogVix [38]

<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

3 0
2 years ago
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A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to d
Nataly_w [17]

Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

R = 0.082atm.L/mol/K

n =?

PV = nRT

n = PV /RT

n = (1.5x12)/(0.082x280)

n = 0.784mol

Next, we convert this mole ( i.e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

Therefore, 29.792g of F2 will react with Xg of NaCl i.e

Xg of NaCl = (29.792 x 117)/38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2

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2 years ago
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asambeis [7]

Answer:

D.

Explanation:

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