How are evidence and conclusions related

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago

A student sets up the following equation to convert a measurement.

dybincka [34]

Answer:

$\frac{1 m}{100 cm}$

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0

2 years ago

The pathway by which carbon is transferred from living organisms to the atmosphere is called:

Phoenix [80]

Answer:

cellular respiration

Explanation:

All exergonic processes produced in the cell, through which substances oxidize and chemical energy is released, are grouped under the name of cellular respiration, but to break down an organic molecule the cells employ, mainly dehydrogenations that can be carried carried out in the presence or absence of atmospheric O2 oxygen. There are therefore two types of breathing: aerobic respiration and anaerobic respiration. The latter also called fermentation.

Aerobic respiration (oxidative phosphorylation)

- Use molecular O2.

- It degrades glucose to CO2 and H2O

- Exergonic

- Recovers about 50% of chemical energy

- Present in most organisms.

- It uses enzymes located in the mitochondria.

3 0

3 years ago

In calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the

BartSMP [9]

Answer: The given statement is TRUE.

Explanation:

An equilibrium reaction is one in which rate of forward reaction is equal to the rate of backward reaction.

Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.

For example for the given equilibrium reaction;

$2H_2O(g)\leftrightharpoons 2H_2(g)+O_2(g)$

$K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Thus the given statement that in calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the factors in the equilibrium expression is True.

7 0

3 years ago

Why don't the praticles in a solid move past one another

Ne4ueva [31]

Answer:

Particles in a solid object are super tight and close to eachother. They do not move past eachother because it is a solid, not a liquid or gas.

Explanation:

3 0

4 years ago

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