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3 years ago

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

Chemistry

1 answer:

ivanzaharov [21]3 years ago

3 0

Answer:

202 g/mol

Explanation:

Let's consider the neutralization between a generic monoprotic acid and KOH.

HA + KOH → KA + H₂O

The moles of KOH that reacted are:

0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol

The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.

1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

0.2688 g/1.33 × 10⁻³ mol = 202 g/mol

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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

These statements describe three different reactions.

Naily [24]

An atom gains an electron from another atom. Hence, option B is the correct answer.

An atom is a particle of matter that uniquely defines a chemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons.

When an atom shares electrons with another atom then it results in the formation of a covalent bond.

Whereas when an atom transfer electrons from one atom to another then it results in the formation of an ionic bond.

When the nucleus of an atom splits then it represents a nuclear fission reaction and energy is released during this process.

Hence, option B is the correct answer.

Learn more about the atom here:

brainly.com/question/1566330

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2 years ago

You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.

Nitella [24]

The question requires us to use the dilution formula $M_iV_i=M_fV_f$ , where $M_i$ and $V_i$ are the stock concentration and volume respectively, then $M_f$ and $V_f$ are the dilute concentration and volume respectively.

a. $C_s_t_o_c_k$ = 1.5 M KCI, $C_d_i_l_u_t_e$ =0.20M KCl, $V_d_i_l_u_t_e$ =250ml KCl

b. $M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml$

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.

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