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Amanda [17]
3 years ago
13

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

ed during the reaction. A cross drawn at the base of each flask became gradually invisible due the formation of this yellow precipitate. The time taken for the cross to become invisible was recorded. A partial record of the experiment is shown. Experimental Record Flask Volume of HCl Volume of Sodium Thiosulfate Volume of Water Time 1 10 mL 10 mL 40 mL 14 seconds 2 10 mL 20 mL 30 mL 3 10 mL 30 mL 20 mL 4 10 mL 40 mL 10 mL Based on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the experimental record. Use complete sentences to explain the trend you predicted. You do not have to determine exact values for time; just describe the trend you would expect (increase or decrease) and why it occurs.
Chemistry
1 answer:
erastovalidia [21]3 years ago
5 0

Answer:

The time taken for the cross mark to disappear decreases steadily down the column.

Explanation:

Now if we look at the data provided, we will discover that the volume of the HCl was held constant while the volume of the thiosulphate was increased steadily and the volume of water decreased steadily.

Recall that a system is more concentrated when it contains less volume of water and more volume of reactants. Hence as the volume of water in the system is being reduced, the concentration of reactants is increased.

It has been established that an increase in the concentration of reactants lead to an increase in the rate of reaction. The disappearance of the cross shows the completion of the reaction between HCl and thiosulphate. The faster or slower the cross disappears, the faster or slower the rate of reaction.

Since increase in concentration of reactants increases the rate of reaction, it is observed that as the volume of the thiosulphate increases (reactant concentration increases) the cross disappears faster (rate of reactant increases). Hence as the volume of thiosulphate increases, it takes a shorter time for the cross to disappear. This implies that the time column in the table (refer to the question) will decrease steadily as the volume of thiosulphate increases.

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Answer

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Since the metals are further to the left on the periodic table, they have low ionization energies and low electron affinities, so they lose electrons relatively easily and gain them with difficulty.  They also have relatively few valence electrons, and can form ions (and thereby satisfy the octet rule) more easily by losing their valence electrons to form positively charged cations.

The main-group metals usually form charges that are the same as their group number:  that is, the Group 1A metals such as sodium and potassium form +1 charges, the Group 2A metals such as magnesium and calcium form 2+ charges, and the Group 3A metals such as aluminum form 3+ charges.

The metals which follow the transition metals (towards the bottom of Groups 4A and 5A) can lose either their outermost s and p electrons, forming charges that are identical to their group number, or they can lose just the p electrons while retaining their two s electrons, forming charges that are the group number minus two.  In other words, tin and lead in Group 4A can form either 4+ or 2+ charges, while bismuth in Group 5A can form either a 5+ or a 3+ charge.

The transition metals usually are capable of forming 2+ charges by losing their valence s electrons, but can also lose electrons from their d orbitals to form other charges.  Most of the transition metals can form more than one possible charge in ionic compounds.

Nonmetals are further to the right on the periodic table, and have high ionization energies and high electron affinities, so they gain electrons relatively easily, and lose them with difficulty.  They also have a larger number of valence electrons, and are already close to having a complete octet of eight electrons.  The nonmetals gain electrons until they have the same number of electrons as the nearest noble gas (Group 8A), forming negatively charged anions which have charges that are the group number minus eight.  That is, the Group 7A nonmetals form 1- charges, the Group 6A nonmetals form 2- charges, and the Group 5A metals form 3- charges.  The Group 8A elements already have eight electrons in their valence shells, and have little tendency to either gain or lose electrons, and do not readily form ionic or molecular compounds.

Ionic compounds are held together in a regular array called a crystal lattice by the attractive forces between the oppositely charged cations and anions.  These attractive forces are very strong, and most ionic compounds therefore have very high melting points.  (For instance, sodium chloride, NaCl, melts at 801°C, while aluminum oxide, Al2O3, melts at 2054°C.)  Ionic compounds are typically hard, rigid, and brittle.  Ionic compounds do not conduct electricity, because the ions are not free to move in the solid phase, but ionic compounds can conduct electricity when they are dissolved in water.

Explanation:

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3 years ago
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4 years ago
A volume measured by a graduated cylinder that was marked in 100 mL
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2 years ago
How many grams of precipitate will be formed when 20.5 mL of 0.800 M
Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0
3 years ago
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