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svp [43]
3 years ago
13

Reaction intermediates differ from activated complexes in that A. they are stable molecules with normal bonds and are frequently

isolated. B. they are molecules with normal bonds rather than partial bonds and can occasionally be isolated. C. they are intermediate structures which have characteristics of both reactants and products. D. they are unstable and can never be isolated. E. all reactions involve reaction intermediates, but not all have activated complexes.
Chemistry
1 answer:
vova2212 [387]3 years ago
3 0

Answer:

they are molecules with normal bonds rather than partial bonds and can occasionally be isolated.

Explanation:

In chemistry, reaction intermediates are species that are formed from reactants and are subsequently being transformed into products as the reaction progresses. In other words,  reaction intermediates are species that do not appear in a balanced reaction equation but occur somewhere along the reaction mechanism of a non-elementary reaction. They are usually short lived species that possess a high amount of energy. They may or may not be isolated.

They are often molecular species with normal bonds unlike activated complexes that are sometimes hypervalent species.

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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
andre [41]

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}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{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*\frac{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*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
Read 2 more answers
How are Ionic and Covalent Bonds are formed with examples ?
Helga [31]

Answer:Comparison of Ionic and Covalent Bonds

In an ionic bond, the atoms are bound together by the electrostatic forces in the attraction between ions of opposite charge. ... For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl. In a covalent bond, the atoms bond by sharing electrons.

     

pls   add  Brainliest

7 0
2 years ago
An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz =106s–1). Calculate the w
kap26 [50]

Answer:

Wavelength=2.899\ m

Explanation:

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength

c is the speed of light having value 2.998\times 10^8\ m/s

Given, Frequency = 103.4 MHz = 103.4\times 10^{6}\ Hz  ( as 1 MHz = 10⁶ Hz)

Thus, Wavelength is:

Wavelength=\frac{c}{Frequency}

Wavelength=\frac{2.998\times 10^8}{103.4\times 10^{6}}\ m

Wavelength=2.899\ m

5 0
3 years ago
When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th
kipiarov [429]

The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

3 0
2 years ago
Read 2 more answers
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