20 Most Recent MPS Multimedia QuickStudy Chemistry for PC Questions & Answers


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Yearning doctors and **** specialists must take and go no less than two semesters of natural science; and before they can even take that course, they should likewise take and pass two semesters of general chemistry, science I and science 2, and for the most part, they should win An or B in these courses to be focused on passage into their preferred scholastic projects.
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MPS Multimedia... | Answered on Oct 25, 2017


i think its 7...

"7.) calcium oxide + diphosphorus decaoxide"

that causes the problem - stop drinking it with empty stomach...

MPS Multimedia... | Answered on Jan 11, 2011


What is the S-F bond energy in SF6 given DHfo for each of the reaction components?

For this problem you must first calculate the change in enthalpy (heat transfer, delta H under standard conditions) for the conversion, S(g) + 6F(g) => SF6(g) OR for the reverse reaction, it really doesn't matter, because the numerical value will be the same, regardless. Once you calculate the heat transferred, you will be able to say that the amount of heat transferred was the amount of potential energy trapped in all of the S-F bonds, all six of them in the molecule. So, to obtain the answer asked for (the energy content on one S-F bond), all you will have to do is divide by 6.

The above is a logical approach, because all six of the S-F bonds are identical. This is so, because, according to VSEPR theory (look it up for more background on that, if you are interested), you can predict that its molecular geometry is "octahedral" with the central sulfur atom having "sp3d2 hybridization."

Here is how you set up the problem:

First write the balanced chemical equation with the given heats of formation (in kJ) written under each of the reaction components:
S(g) + 6F(g) => SF6(g)
275 6(80.) -1100.

Note: I am assuming that each of the above quanties is good (i.e., known) to at least the unit's place; that is, + or - 1 kJ. This reasonable assumption allows me to unambiguously indicate the number of sig figs in each quantity - an important consideration for proper rounding off of the final answer.

Recall that the sum of the product values minus the sum of the reactant values, each component multiplied by its corresponding coefficient will give the net enthalpy ("reaction enthalpy") of the reaction as written. In this case, there being only one product, the reaction enthalpy is:
-1100 - (480 + 275) = - 1855 kJ. From this, we can see that as S and six Fs are combined, 1,855 kJ of heat are released into the surroundings (that is, an exothermic reaction). The amount of heat released informs you of the combined bond energy of ALL six S-F bonds.

A good rule to remember: As bonds are formed, energy is always released (an exothermic process). As bonds are broken (as in the reverse reaction), the same amount of energy is being absorbed (an endothermic process).

Therefore, in conclusion, one S-F bond has a bond energy of 309.17, which is more properly rounded off to 309 kJ.

Recall that the rule for rounding off when adding or subtracting is to make sure that the final answer has the same precision as the values used to calculate it. Since each given value was good only only to the last unit

MPS Multimedia... | Answered on Jan 05, 2011


Most carbon compounds that are used for fuels belong to the category of "organic" compounds called "hydrocarbons (HCs)," which, historically, at least, are plentiful and harvested from ancient petroleum deposits by the big oil companies - the raw oil being easily fractionated at processing plants to produce a great variety of different organic compounds from the most volatile (gases, such as CH4, methane; C2H6, ethane; C3H8, propane, etc) to an extensive series of progressively higher boiling liquids (C4H10, butane; C5H12, pentane; C6H14, hexane; C7H16, heptane; C8H18, octane; etc. Our gasoline is usually a complex mixture of these hydrocarbons, in the range between 3 and 12 carbons.

They are a traditional favorite for fuels because they can be stored in condensed form and transported long distances via truck or a complex underground system of pipes; and because they burn cleanly in the presence of sufficient oxygen to give only carbon dioxide (CO2, not classified as an organic compound) and water (H2O), each of which may be trapped and recycled in industrial processes. Another big plus for the use of HCs is that all of their reactions in the presence of oxygen are "exothermic"; that is, as they burn heat is always produced, which can be use to heat our homes and provide the energy needed to run our machines, such as our cars and trucks and jets.

MPS Multimedia... | Answered on Jan 05, 2011


Hello, I can help you if you include more data in your question.

As an organic compound, lidocaine is insoluble in water, so I am assuming that either some combination of water and alcohol (ethanol), or a pure organic solvent like ethanol or diethyl ether is being used.

If lidocaine were soluble in water (which it's not), you would have been able to solve your problem as indicated below. Maybe you would be able to modify the indicated solution by taking into consideration the density of the organic solvent.

Percentage (%) by mass is one of the common ways to express the concentrations of solutions.
Since % means the number of grams of solute in 100 grams of solution (solute + solvent), a 1.5% solution of any solute refers to 1.5 grams of it in 100 grams of solution.

Assuming we are referring to an aqueous solution (that is, where water is the solvent), this reasonably approximates to 1.5 grams of lidocaine per 100 mL solution. We can assume this result, because we have a dilute watery solution, which probably has the density close to the density of pure water, which is 1.0 gram per 1.0 mL at commonly used temperatures, such as average room temperature.

So, 1.5 g lidocaine per 100 mL solution can easily be scaled down by one-tenth to give:
0.15 g lidocaine per 10 mL solution.

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MPS Multimedia... | Answered on Jan 04, 2011


Access Google Australia: http://www.google.com.au
and enter "australian university petroleum chemistry"
and do a search.

MPS Multimedia... | Answered on Sep 27, 2010


0 degree K is equal to -273.15 degree C

MPS Multimedia... | Answered on Jun 26, 2010


science is the well organized intelligence about every thing .

MPS Multimedia... | Answered on Jun 17, 2010


In fact there are 2 different conservation laws: conservation of mass and conservation of energy. Of course in a chemical or any other physical process these 2 laws will hold!
Conservation of mass: the mass of a closed (isolated system) system will remain constant over time.
Conservation of energy: total amount of energy in an closed system remains constant over time
Furthermore if we look into the relativity theory, it states that in fact mass and energy are two names for the same thing (the equivalence principle of mass and energy), neither one "appears" without the other.
Example: an iron nail left in a cup of water until it turns entirely to rust. A precise scale will register a change in mass. The energy will appear as heat, which will be lost to the environment so because of the conservation law you will see a loss in the mass of the system.

MPS Multimedia... | Answered on Apr 20, 2010


Hi,
The law of conservation of mass and energy states that mass and energy can neighther be created nor destroyed however it can be changed into one form to another.

MPS Multimedia... | Answered on Mar 11, 2010


Protons have a +1 charge. This is the same magnitude of an electron but an electron has a −1 charge. Neutrons have a 0 or no charge.

The following Wiki Article explains the process of Proton decay

Main article: Proton decay Protons are observed to be stable and their theoretical minimum half-life is 1036 years. Grand unified theories generally predict that proton decay should take place, although experiments so far have only resulted in a lower limit of 1035 years for the proton's lifetime. In other words, proton decay has never been witnessed and the experimental lower bound on the mean proton lifetime (2.1×1029 years) is put by the Sudbury Neutrino Observatory.[5]

However, protons are known to transform into neutrons through the process of electron capture. "When a high energy-proton collides with an atom, it causes the ejection of an electron from the outer layer of the atom."[6]:125 This process does not occur spontaneously but only when energy is supplied. The equation is:
1a11b43084f71df3683387ac7281c27b.png where
p is a proton, e is an electron, n is a neutron, and νe is an electron neutrino The process is reversible: neutrons can convert back to protons through beta decay, a common form of radioactive decay. In fact, a free neutron decays this way with a mean lifetime of about 15 minutes.

MPS Multimedia... | Answered on Oct 16, 2009


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