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3 years ago

What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?

1 answer:

4 0

If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x

⇒ x = $\frac{250 ml * 1.25mol}{1000 ml}$

= 0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole × molar mass

= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol

= 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is 18.75 g

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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

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The estimate obtained from a sample of which size is likely to be closest to

zubka84 [21]

The answer is C, 85.

8 0

3 years ago

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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

When NADH donates two electrons to ubiquinone during respiration, ubiquinone isa. reduced.b. oxidized.c. phosphorylated.d. aerob

olga nikolaevna [1]

Reduced ... more electrons gain is a reduction

5 0

3 years ago

When a soda can is dropped, it should not be immediately opened. Why?

Kipish [7]

The pressure inside a coke bottle is really high. This helps keep the soda carbonated. That is, the additional pressure at the surface of the liquid inside the bottle forces the bubbles to stay dissolved within the soda. When the coke is opened, there is suddenly a great pressure differential. The initial loud hiss that is heard is this pressure differential equalizing itself. All of the additional pressure found within the bottle pushes gas out of the bottle until the pressure inside the bottle is the same as the pressure outside the bottle. However, once this occurs, the pressure inside the bottle is much lower and the gas bubbles that had previously been dissolved into the soda have nothing holding them in the liquid anymore so they start rising out of the liquid. As they reach the surface, they pop and force small explosions of soda. These explosions are the source of the popping and hissing that continues while the soda is opened to the outside air. Of course, after a while, the soda will become "flat" when the only gas left dissolved in the liquid will be the gas that is held back by the relatively weak atmospheric pressure.

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3 years ago