Why is it easier to demolish a building than it was to build it

Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.

Harlamova29_29 [7]

Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-(aq)
OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-(aq)

Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+]/[HOCl] ............. (1)
∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8) ............. (2)

and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3)
Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8)
we know that, [OH-][H+] = 10-14

∴K = 3.3 * 10-7

Thus, correct answer is e i.e none of these

6 0

3 years ago

Read 2 more answers

Which choice represents a pair of resonance structures? Note that lone pairs have been omitted for clarity. View Available Hint(

stepladder [879]

Answer:

A. O=C=O and O≡C−O

Explanation:

Resonance:

When the electron distribution on the molecule become uneven like one molecule have more electron compare to other.Resonance occurs due to overlap of the orbitals.When electron flow through pi system then resonance occurs.

So the option A is correct.

A. O=C=O and O≡C−O

6 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the iden

Sindrei [870]

Answer:

HCl

Explanation:

Choices:

CO: 28.01g/mol

NO₂: 46g/mol

CH₄: 16.04g/mol

HCl: 36.4g/mol

CO₂: 44.01g/mol

It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:

PV = nRT

Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).

Replacing:

PV = nRT

PV / RT = n

0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n

8.786x10⁻³ = moles of the gas

As the mass of the gas is 0.320g; its molar mass is:

0.320g / 8.786x10⁻³moles = 36.4 g/mol

Based in the group of answer choices, the identity of the gas is:

4 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago