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Serga [27]
3 years ago
6

If the relative activities of two metals are known which metal is more easily oxidized

Chemistry
1 answer:
Ksivusya [100]3 years ago
6 0
Answer: The metal that has a greater reactivity is more easily oxidized.

Explanation:

Oxidation is when the elements lose electrons and increase their oxidation state.

The metals tend to react by losing electrons and form the corresponding cation.

For expample, sodium (an alkalyne metal) loses one elecron and form the cation Na¹⁺ , then this cation combine with an anion and form compounds like NaCl, NaOH. The same do the other alkalyne metals.

Magnesium (an alkalyne earth metal) loses two electrons and form the cation Mg²⁺, then it combines with some anions to form compounds, like MgSO₄, Mg(OH)₂.

So, the easier the metal gets oxidized the greater its reactivity.




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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
Can light bend around corners to reach an object
Sholpan [36]

Answer: Yes, light can bend around corners. In fact, light always bends around corners to some extent.

Explanation:This is a basic property of light and all other waves. ... The ability of light to bend around corners is also known as "diffraction".

7 0
3 years ago
Read 2 more answers
What is the Kb expression for aniline (C6H5NH2)?
Dimas [21]
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]

hope this helps:)

</span>
4 0
3 years ago
Read 2 more answers
Potassium impart purple colour but beryllium do not impart any colour to the flame.Why?​
BaLLatris [955]

Answer:

<h2>The electrons in beryllium and magnesium are too strongly bound to get excited by flame. Hence, these elements do not impart any color.</h2>
6 0
3 years ago
ammonia gas is used as refrigerant 0.474 atm. Pressure is required to change 2000 cm3 sample of ammonia initially at 1.0 atm to
Andreas93 [3]

Answer: The pressure required is 0.474 atm

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}V}   (At constant temperature and number of moles)

The equation is,

{P_1V_1}={P_2V_2}

where,

P_1 = initial pressure of gas = 1.0 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 2000cm^3

V_2 = final volume of gas = 4.22dm^3=4220cm^3    (1dm^3=1000cm^3)

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

{1.0\times 2000}={P_2\times 4200}

P_2=0.474atm

The pressure required is 0.474 atm

5 0
2 years ago
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