Common Laboratory Equipment

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Common Mistakes when Pitching

Ever wonder what the most common miss/hit golfswings are when pitching a golf ball? It's when you top it, when you skull it, or when you hit it real, real thin and it runs over the green. I'm going to tell you why that happens.

A little review of the setup: ball in the middle of your stance, your weight left and handle forward. But here's the first mistake most of you make.

When the club starts back, you start back. Now 70% of your weight isn't on your left leg. And if you do that, it's very difficult to get the club underneath the golf ball.

So, the most common golfswing is when you slide backwards and you hit the ground behind it, or you top it or skull it.

Now, here's the next most common mistake that I've seen over the years. Everybody thinks that to hit a great pitch shot you have to have this beautiful, high followthrough.

As a matter of fact, that is not true for pitching the ball up in the air.

It's a very pretty golfswing, but so many of you have been sold on the idea that you have to follow through real high.

To stop from skulling or topping the ball, weight in the middle, handle left, weight on your left side, swing the arms up and swing the arms down in the downswing.Up/down makes the golf ball go up.

Now, many golfers hit it fat. This is caused by the club being to vertical. If you go too vertical, you'll wind up chopping the floor.

After you get all set up, make sure you're not leaning too far over on the left side. A little weight on your left side, but the trick is to swing the golf club up on the inside. And that will give the golf club a little better angle into the back of the ball.

Try these golfswings the next time you're out. I promise you it'll help you.

Thanks.Bobby Eldridge is the Head Instructor for the PurePoint Golf Academy where he teaches "The Simple Golf Swing" theory. You can check out PurePoint Golf instructional DVDs athttp://www.golfswingguru.com/index15.htm

Article Source: http://www.simplysearch4it.com/article/52157.html

Common Laboratory Equipment

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Common Scientific Lab Equipment

Frequently Asked Questions...

answer question below!?

We've been assigned a lab in which we have to determine the percentages by mass of two solids in a mixture. The two solids in the mixture are sodium chloride and sodium bicarbonate. Available to us are common laboratory equipment, as well as 1 M hydrochloric acid and 1 M sulfuric acid. In addition, we also have the pure solids themselves.

I've tried to do some chemical equations (specifically, taking each solid in the mixture and then reacting them with either HCl or H2SO4), but I'm still relatively clueless as to how to isolate each component within the mixture.

Also, is bicarbonate similar to carbonate in that it reacts with an acid to produce carbon dioxide, water, and a salt?

Thanks for any help you can give me.


Best Answer...

Answer:

The easiest thing to do would be to weigh a sample of the mixed chemicals, then dissolve them in water. Next, add sufficient HCl to bring the pH of the solution to around pH 1-2. This will convert all the sodium bicarbonate to sodium chloride via the reaction:

HCO3- + H+ -> H2O + CO2(g)

The NaCl will not react with the HCl, so it's just a "bystander" in this reaction.

Now you have a solution of NaCl, perhaps with a small amount of dissolved carbonic acid (H2CO3). Evaporate this solution to dryness, and reweigh the solids, which at this point will be all NaCl.

The mass of the sample before reacting it with HCl is the sum of the masses of NaCl (m_NaCl) and NaHCO3 (m_NaHCO3):

M_init = m_NaCl + m_NaHCO3

Each mole of NaHCO3 that reacts forms one mole of NaCl, so, the mass of NaCl that is produced by reaction with HCl is:

(m_NaHCO3/MW_NaHCO3)*MW_NaCl

where MW_NaHCO3 and MW_NaCl are the molecular masses of sodium bicarbonate (84.007 gm/mol) and socium chloride (58.442 gm/mol), respectively.

The mass of the dried solids after the reaction is then:

M2 = m_NaCl + (m_NaHCO3/MW_NaHCO3)*MW_NaCl
M2 = m_NaCl + (84.007/58.442)*m_NaHCO3
M2 = n_NaCl + 1.437*m_NaHCO3

Subtract the mass of the sample after reaction from the mass before reaction:

M2 - M1 = (n_NaCl + 1.437*m_NaHCO3) - (m_NaCl + m_NaHCO3)

M2 - M1 = 0.437*m_NaHCO3

(M2-M1)/0.437 = m_NaHCO3

This gives you the mass of sodium carbonate that was originally in the sample. The percentage by mass of sodium carbonate is then simply given by:

100 * m_NaHCO3 / M1

The percentage by mass of NaCl is then 100% minus the wt. percent NaHCO3. Similarly, the mass of NaCl is just M1 minus the mass of NaHCO3.

There are many other ways one might do this, but this seemed to be the simplest. One just needs to be careful to recover all the dried sample after the reaction (probably best to do the evaporation in a pre-weighed beaker, so the solids don't need to be transferred. One also needs to be sure that the solids are completely dry.

You could use the sulfuric acid to do the same thing, but if you add excess acid, sulfuric acid is harder to evaporate to dryness. NaSO4 is also a more complicated salt than NaCl (see source below), forming a stable hydrate at low temperatures.

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