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3 years ago

Which of the following is a property of matter?

Chemistry

1 answer:

lana66690 [7]3 years ago

5 0

Explanation:

Matter also exhibits physical properties. Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.

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If a strong base, like NaOH, is added to a strong acid, like HCl, the pH of the resultant solution will _________. A.) Go up B.)

MakcuM [25]

Answer:

B.) Go down

Explanation:

Hello,

In this case, a strong base like NaOH has a high pH based on the scale wherein 7 is neutral, above 7 is basic and below 7 is acid. In such a way, by adding an acid, having a low pH, once it is added, the pH will go down until the equivalence point, which for strong base and strong acid should be 7.

Best regards.

7 0

3 years ago

Read 2 more answers

When naming acids, the prefix hydro- is used when the name of the acid anion ends in _____?

Natali5045456 [20]

<span>When naming acids, the prefix hydro- is used when the name of the acid anion ends in "-ide". Simple acids, known as binary acids, have only one anion and one hydrogen. These anions usually have the ending -ide. For example, hydrogen chloride becomes hydrochloric acid.</span>

8 0

3 years ago

Identify the following as an example of a physical property or a chemical property. Silver tarnishes when it comes into contact

Assoli18 [71]

The answer would be A) Physical Change. When the silver tarnishes it is still remnants of silver. If it were B, the silver would change into a different substance.

6 0

3 years ago

Read 2 more answers

Freon-12 CCl2F2, is prepared from CCl4 by reaction with HF. The other product of this reaction is HCl. Outline the steps needed

LekaFEV [45]

Answer:

percent yield = 40.6 %

Explanation:

The question asks to determine the percent yield, which can be defined as:

percent yield = $\frac{actual yield}{theoretical yield} *100$

where the actual yield is how much product was obtained, in this case 12.5 g of CCl₂F₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically.

So we know already the actual yield, we need to <em>calculate the theoretical yield.</em>

First we need to <em>write the reaction chemical equation</em>:

CCl₄ + HF → CCl₂F₂ + HCl

and <em>balance the equation</em>:

CCl₄ + 2 HF → CCl₂F₂ + 2 HCl

In the equation we can see that <em>for every mol of CCl₄ we should get 1 mol of CCl₂F₂</em> (molar ratio 1:1). So if we <u>calculate the moles of CCl₄</u> in the given 39.2 g of CCl₄ we could know how many moles of CCl₂F₂ (assuming HF is in excess).

Moles of CCl₄ = mass CCl₄ / molar mass CCl₄
Molar Mass CCl₄ = 12.011 + 4 * 35.45 = 153.811 g/mol
Moles of CCl₄ = 39.2 g / 153.811 g/mol = 0.2549 moles

From the molar ratio we know:

Moles of CCl₂F₂ = moles of CCl₄ = 0.2549 moles

Now we need to <u>convert these moles into grams</u> to get the theoretical yield of CCl₂F₂ in grams:

mass CCl₂F₂ = moles CCl₂F₂ * molar mass CCl₂F₂
Molar Mass CCl₂F₂ = 12.011 + 2 * 35.45 + 2 * 18.998 = 120.907 g/mol
Mass CCl₂F₂ = 0.2549 moles * 120.907 g/mol = 30.81 g

Theoretical yield CCl₂F₂ = 30.81 g

Percent yield = (12.5 g / 30.81 g) * 100 = 40.6 %

8 0

3 years ago

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}$ .

8 0

3 years ago