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densk [106]
1 year ago
6

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) P₄(s) → HPO₃²⁻(aq) + PH₃(g) [acidic]

Chemistry
1 answer:
m_a_m_a [10]1 year ago
5 0

The balanced chemical equation is :

5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)

Here the oxidation number of P changed from 0 to -3 in PH₃ and increases  from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.

Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

brainly.com/question/13182308

#SPJ4

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Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH–] is 1 x 10–11 mol/L. Is this solution acidi
valina [46]
(H+)(OH-) = Kw
kw=  1 x10^-14
OH-=  1   x10 ^-11
(H+)=  KW / OH-

concentration   of H+  = (1x10^-14) /.(1  x 10 ^-11)   =  1  x10  ^-3

Ph=  -log (H+)

PH=-log (  1  x  10  ^-3)  =  3  therefore  the    solution  is  acidic  since  the  PH   less than  7

7 0
3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
When 2.16g of H2 reacts with excess O2 by the following equation, 258 kJ of heat are released. What is the change of enthalpy as
alekssr [168]

Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

8 0
2 years ago
When 1.025 g of naphthalene (c10h8) burns in a bomb calorimeter, the temperature rises from 24.25°c to 32.33°c. find δerxn for t
Goryan [66]

Answer : The \Delta E for the combustion of naphthalene is 5161.25KJ/mole

Solution : Given,

Mass of naphthalene = 1.025 g

Initial temperature = 24.25^oC

Final temperature = 32.33^oC

Specific heat capacity of calorimeter = 5.11KJ/^oC

Molar mass of naphthalene = 128 g/mole

First, we have to calculate the heat absorbed, q

Formula used :

q=c\times \Delta T=c\times (T_{final}-T_{initial})

Now put all the given values in this formula, we get

q=(5.11KJ/^oC)\times (32.33^oCT-24.25^oC)=41.29KJ

Now we have to calculate the moles of naphthalene.

Moles of C_{10}H_{8} = \frac{\text{ Mass of }C_{10}H_{8}}{\text{ Molar mass of }C_{10}H_{8}}=\frac{1.025g}{128g/mole}=0.0080moles

Now we have to calculate the \Delta E for combustion of naphthalene.

\Delta E=\frac{q}{n}

where,

q = heat absorbed

n = number of moles

Now put all the values in this formula, we get

\Delta E=\frac{41.29KJ}{0.0080moles}=5161.25KJ/mole

Therefore, the \Delta E for the combustion of naphthalene is 5161.25KJ/mole

6 0
2 years ago
When is the BEST time to be treated for an STI?
allochka39001 [22]
The best answer for this question is <span>d. A person should seek treatment the first moment he or she suspects having an STI.</span><span>

The moment a person suspects that he/she has an STI, that person must immediately have him/her self checked by a doctor. The earlier an STI can be diagnosed, the earlier it can be treated. A person does not have to wait for a symptom to appear before treating the infection.</span>
3 0
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