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CaHeK987 [17]
3 years ago
12

Students find the pH of Substance A to be 2.1 and Substance B to be 4.4. They are told Substance C is less acidic than Substance

A but more acidic than Substance B. What is the possible pH RANGE of Substance C?
pls don’t give me a link
Chemistry
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

the range should be 2.2 to 4.3

Explanation:

I think so because the numbers at the left side of the scale from 1 are more acidic so as it increases it's still acidic but lesser so 1 is more acidic than 2 so I used 2.2 as the beginning of the range because it's less acidic than A even though its a greater number and 4.3 is lesser than 4.4 but its still greater on the scale. frankly speaking I don't feel so correct because it's in decimal so try and compare facts thank you

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Answer:

The answer to your question is  CH₄

Explanation:

Process

1.- Write the possible chemical reaction

                   CxHy + O₂   ⇒    CO₂  +   H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

                         44 g of CO₂ ------------------- 12 g of carbon

                         93.5 g of CO₂ ----------------  x

                          x = (93.5 x 12) / 44

                          x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

                         18 g of water -------------   2 g of hydrogen

                          76.5 g of water --------    x g of hydrogen

                           x = (76.5 x 2) / 18

                           x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

                           12g of carbon ------------- 1 mol

                           25.5g of carbon ---------- x

                             x = (25.5 x 1) / 12

                              x = 2.12 moles of carbon

                            1 g of hydrogen ----------- 1 mol

                            8.5 g of hydrogen -------- x

                             x = (8.5 x 1) / 1

                             x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

                                                    CH₄                                        

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Explanation :

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=(^oF-32)\times \frac{5}{9}

^oC=(75-32)\times \frac{5}{9}=24^oC

The conversion used for the temperature from degree Celsius to Kelvin is:

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K=273+24=297K

n = number of moles

m = mass of air

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\rho = density of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above formula, we get:

\rho=\frac{PM}{RT}

\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}

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Conversion used :

1g/L=0.0624lbm/ft^3

So,

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The density of air in lbm/ft^3 is, 0.0743lbm/ft^3

Now we have to calculate density in kg/m^3.

Conversion used :

1g/L=1kg/m^3

So,

1.19g/L=1.19kg/m^3

The density of air in kg/m^3 is, 1.19kg/m^3

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