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3 years ago

14

A substance undergoes a change. Which of the following indicates that the change was a chemical change? The substance changed sh

ape. New molecules formed. The substance froze. The substance changed size.

2 answers:

irina1246 [14]3 years ago

8 0

Answer: New molecules formed.

Explanation:

A physical change is defined as the change in which shape, size will be altered. No new substance gets formed in these reactions.

A chemical change is defined as the change in which chemical composition is altered. A new substance is formed in these reactions.

The substance changed shape is a physical change.

The substance froze means the state is changed and is a physical change.

The substance changed size is a physical change.

Thus when new molecules are formed, it is a chemical change.

MariettaO [177]3 years ago

3 0

In order to form new molecules, a chemical reaction would have to occur which means the change would be a chemical change.

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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

Has 3 valence electrons and 4 energy levels ?

AVprozaik [17]

You are looking for an element in the fourth period and a transition metal. As most have about 3 valence electrons. Gallium and Scandium work perfectly.

8 0

3 years ago

Read 2 more answers

Who discovered phosphate?

iris [78.8K]

Answer:

Hennig Brand

Explanation:

4 0

3 years ago

(a) Write a briefexperimental procedure for the chromic acid test.Include all observations such ascolor change, precipitation, e

mel-nik [20]

Answer:

See explanation

Explanation:

a) -About 1-2 drops of the unknown is dissolved in approximately 1 mL of analytical grade acetone standard solution. The test solution is added and shaken thoroughly. If a positive test is not immediately observed, the set should be allowed to stand for 1-2 minutes. A positive result means the appearance of a green color.

b) The functional groups that can be distinguished by the test are alcohols and aldehydes.

Aldehydes give a positive test to chromic acid but ketones do not.

Primary and secondary alcohols give a positive test to chromic acid but tertiary alcohols do not.

c) Chromic Acid Test involves Cr in the +6 oxidation state. A positive test implies the reduction of orange Cr^6+ to green chromium Cr^3+.

d) The compounds were not shown but this image attached from lumen learning summarizes the reaction mechanism of chromic acid test.

3 0

3 years ago

A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure su

Tom [10]

Answer:

$V_2 = 4.87 * 10^3$

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

$PV = nRT$

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

$\frac{PV}{PT} = \frac{nRT}{PT}$

$\frac{V}{T} = \frac{nR}{P}$

Represent $\frac{nR}{P}$ with k

$\frac{V}{T} = k$

$k = \frac{V_1}{T_1} = \frac{V_2}{T_2}$

At this point, we can solve for the required parameter using the following;

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$\frac{4.5 * 10^3}{298} = \frac{V_2}{323}$

Multiply both sides by 323

$323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323$

$323 * \frac{4.5 * 10^3}{298} = V_2$

$V_2 = 323 * \frac{4.5 * 10^3}{298}$

$V_2 = \frac{323 * 4.5 * 10^3}{298}$

$V_2 = \frac{1453.5 * 10^3}{298}$

$V_2 = 4.87 * 10^3$

Hence, the final volume at 50C is $V_2 = 4.87 * 10^3$

7 0

3 years ago