Classify each of the following chemical reactions as a synthesis decomposition and single displacement or double displacement reaction

Answers

Answer 1

Answer:

1. Synthesis reaction : there is only 1 product formed from 2 or more reactant

E.g:

$H_2(g)+N_2(g)\text{ }\Rightarrow2NH_3(g)\text{ }$

2. Decomposition : reaction that occurs in presence of UV light and only 1 reactant that decomposes into 2 or more products.

E.g:

$CH_3Br(g)+UV_(light)\Rightarrow CH_3(g)\text{ + Br (g)}$

3. Single displacement :reaction that occurs when 1 reactant displaces other reactant from its compound:

E.g:

$Zn(s)+CoCl_2(aq)\text{ }\Rightarrow ZnCl_2(aq)\text{ + Co(s)}$

4. Double displacement :reaction that occurs when both reactant displaces each other.

E.g :

$K_2S(aq)+Co(NO_3)_2\Rightarrow2KNO_3(aq)\text{ + }CoS(s)\text{ }$

Related Questions

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated
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Ne ( g ) effuses at a rate that is ______ times that of Cl 2 ( g ) under the same conditions.

Theoretical yield
2.05 g salicylic acid x (180g aspirin/1 mol) x (1 mol/138 g salicylic acid)

Answers

The question is incomplete; part of the data required in the question are shown:

Theoretical Yield: 2.05 g salicylic acid x (180g aspirin/1 mol) x (1 mol/138 g salicylic acid) 2. Mass of filter paper 2.56 g 3. Mass of filter paper and aspirin 5.42 g 4. Mass of aspirin (3-2) g. Percent Yield [(4)/(1)] x 100

Answer:

107%

Explanation:

We can calculate the theoretical yield as shown;

2.05g salicylic acid × 180g aspirin/1mol × 1 mol/138g of salicylic acid

Theoretical yield= 2.67 g of aspirin

Actual yield of aspirin is obtained from the experimental data;

Mass aspirin + filter paper= 5.42 g

Mass of filter paper= 2.56 g

Mass of aspirin= 5.42 g -2.56 g = 2.86 g

Hence actual yield of aspirin = 2.86 g

Percentage yield = actual yield/theoretical yield × 100

Percentage yield = 2.86/2.67 ×100 = 107%

This is how osmium appears in the periodic table.A purple box has O s at the center and 76 above. Below it says osmium and below that 190.23.Rounded to the nearest whole number, how many neutrons, on average, are in an atom of osmium?

76
114
190
266

Answers

Answer:

it B 114 on edge 2020

Explanation:

UwU

Answer:

it B 114

Explanation:

Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Where would you expect to find the 14C label when alpha-ketoglutarate is converted to succinate?

Answers

Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive $^(14)C$ while none of succinate would contain $^(14)C$ , and all of carbon dioxide released would contain $^(14)C$ .

Explanation:

When oxaloacetate in the citric acid cycle is labeled with $^(14)C$ in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the $^(14)C$ ) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. $CO_2$ released had all the radioactivity.

What is the purpose of a catalyst?O A. To change the potential energy of the reactants
O B. To lower the activation energy of a reaction
O C. To increase the kinetic energy of the reactants
O D. To shift the equilibrium position of a reaction

Answers

B. To lower the activation energy of a reaction

Answer:

To lower the activation energy of a reaction

Explanation:

i just took the test and got it right ...... i hope this helps :)

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

$m_(PbI_2)=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2$

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Which two structures will provide a positive identification of a plant cell under a microscope? A.) Lysosomes, cell wall. B.) large central vacuole, cell wall. C.) large central vacuole ribosomes. D.)nucleoid, chloroplasts.

Answers

The Right Answer Is D.) Nucleoid chloroplasts.

the answer is B large central vacuole and cell wall. they are the easiest/biggest things to see under a microscope to identify a plant cell

Other Questions

Four beakers containing potassium nitrate dissolved in water are allowed to evaporate to dryness. Beakers 1 through 4 contain 2.3, 1.91, 5.985, and 0.52 g of dry potassium nitrate respectively. How many moles of potassium nitrate were recovered after the water evaporated?

HELP ME PLEASE I will mark brainliest

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Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing that the graviatational potential energy of falling water per second = mass (kg) x height (m) x gravity (9.8 m/s2), what is the power of Niagara Falls? How many 15 W LED light bulbs could it power?