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astraxan [27]
3 years ago
15

What is the pH of a 0.1 M acid solution? a) 0 b) -1 c) 1

Chemistry
2 answers:
Alisiya [41]3 years ago
8 0

Answer:

C = 1

Explanation:

For a strong acid, equation of dissociation reaction;

HX(aq) + H2O(l) —> X- + H3O+

For a strong acid, the equilibrium reaction lies strongly to the right.

pH = -log10[H3O+] = -log10[0.1] = -(-1) = 1.

Naddik [55]3 years ago
7 0
Answer : Option C) 1

Explanation : Using the formula for finding pH of strong acid,

pH = - log_{10} [H^{+}] = -log_{10} [0.1]
= -log_{10} 10^{-1} = - (-1) = 1.

This calculation shows how to find the pH of any acid if the concentration is given.
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Answer:

The possible valances can be determined by electron configuration and electron negativity

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Explanation:

All atoms strive for stability. The optima electron configuration is the electron configuration of the VIII A family or inert gases.

Look at the electron configuration of the nonmetal and how many more electrons the nonmetal needs to achieve the stable electron configuration of the inert gases. Non metals tend to be negative in nature and gain electrons. ( They are oxidizing agents)

For example Florine atomic number 9 needs one more electron to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Flowrine has a valance of -1

Oxygen atomic number 8 needs two more electrons to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Oxygen has a valance charge of -2.

Non metals with a low electron negativity will lose electrons when reacting with another non metal that has a higher electron negativity. When the non metal forms an ion it is necessary to look at the electron structure to determine how many electrons the element can lose to gain stability.

For example Chlorine which is normally -1 like Florine when it combines with oxygen can be +1, +3, + 5 or +7. It can lose its one unpaired electron, or combinations of the unpaired electron and sets of the three pairs of electrons.

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The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p
Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

    Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g  

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}

(ii) Solve the relationship

 Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

    56x + 18(7.5 - x) = 267.3

      56x + 135 - 18x = 267.3

                        38x = 132.3

                            x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

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