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3 years ago

15

Chemical A and Chemical B react in an exothermic reaction. What can be known about what will happen when Chemical A and Chemical

B are mixed together?
Please help. I'm confused.

1 answer:

Fynjy0 [20]3 years ago

8 0

In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.

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The validity of scientific theories and laws depends on evidence. Which of the following statements of evidence supports a law r

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d the amount of matter in a closed system is the same at the start of a reaction as at the end of the reaction

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Small quantites of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.

ira [324]

<u>Answer:</u> The mass or zinc reacted is 0.624 grams.

<u>Explanation:</u>

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm (Conversion factor: 1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

$p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm$

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = $30^oC=[30+273]K=303K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol$

The chemical equation for the reaction of zinc and hydrochloric acid follows:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, $9.55\times 10^{-3}mol$ of hydrogen gas is produced from = $\frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol$ of zinc metal

To calculate the mass of zinc metal, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of zinc = 65.38 g/mol

Moles of zinc = $9.55\times 10^{-3}$ moles

Putting values in above equation, we get:

$9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g$

Hence, the mass or zinc reacted is 0.624 grams.

4 0

3 years ago

Two voltaic cells are to be joined so that one will run the other as an electrolytic cell. In the first cell, one half-cell has

vladimir1956 [14]

Au/Cr cell will run as a voltaic cell and Co/Zn cell will run as an electrolytic cell.

<h3>What is an electrolytic cell?</h3>

An electrolytic cell that uses electrical energy to enable a non-spontaneous redox reaction is known as an electrolytic cell.

Certain chemicals can be electrolyzed using electrolytic cells, which are electrochemical cells.

For example, water can be electrolyzed to create gaseous oxygen and gaseous hydrogen with the use of an electrolytic cell.

The following are the three essential parts of electrolytic cells:

Cathode (which is negatively charged for electrolytic cells)

Anode (which is positively charged for electrolytic cells)

Electrolyte

The reactions for each half cell is given by

$Au^{3+}(aq) + 3e^- \rightleftharpoons Au(s)$ ; E° = 1.50 V

$Cr^{3+}(aq) + 3e^- \rightleftharpoons Cr(s)$ ; E° = –0.74 V

$Co^{2+}(aq) + 2e^- \rightleftharpoons Co(s)$ ; E° = –0.28 V

$Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s)$ ; E° = –0.76 V

When the Au/Cr cell will run as a voltaic cell and Co/Zn cell will run as an electrolytic cell, the a difference in the voltages is produced

$E^o_{series} = E^o_{Au/Cr} - E^o_{Co/Zn}$ = 2.24 V – 0.48 V = 1.76 V

Learn more about electrolytic cell:

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1 year ago

A sample of oxygen gas has a volume of 45.8 mL when its pressure is 0.492 atm. What

White raven [17]

Answer:

<h2> a. 23.6</h2>

Explanation:

$P_1=0.492\:atm\\V_1 = 45.8\:ml\\V_2 = ?\\P_1 = 0.954\:atm\\\\Boyle's Law = P_1V_1 = P_2V_2\\\\Lets \:make\: V \:subject\:of\:the\:formula\\\\\frac{P_1V_1}{P_2} = \frac{P_2V_2}{P_2} \\ \\V_2 =\frac{P_1V_1}{P_2} \\V_2 = \frac{0.492\times45.8}{0.954}\\ \\V_2 = \frac{22.5336}{0.954} \\\\V_2 = 23.620\\\\V_2 = 23.6$

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