Question about ATDEC Telehook Fixed Wall Mount for Flat Panels 20 to 37in Silver

I can definitely tell you who can help you fix it, please give me your zip code and I will get right back to you with the information.

Thanks
Tim

Posted on Oct 06, 2010

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Posted on Jan 02, 2017

Mass of 1 mole of CaO (from periodic table)= 40.078g + 15.999g = 56.077g/mole. Thus 0.045mole*56.077 g/mole = 2.523 g or 2.5 g using significant figures.

Dec 11, 2014 | Cars & Trucks

Number of moles [H+] in first solution =concentration of [H+] in mol/L times the number of liters. **[C1]=10^(-pH)=10^(-3) mol /L**

**Number of moles **N1=10^(-3) mol/L * 1L= **10^(-3) mol [H+]**

Concentration [C2] of second solution

[C2]= 10^(-5.5)=3.16227766*10^(-6) mol/L of [H+]

**Number of moles **of H+ in second solution

**N2**=[C2]*10 L =**31.6227766*10^(-6) mol of H+**

Total number of moles H+ in the new solution

**N=10^(-3) +31.6227766*10^(-6) =1.031622777*10^(-3) mol**

Concentration of H+ in new solution

[C]=**1.031622777*10^(-3) mol** / (1+10) L=9.358388878 *10^(-5) mol/L

Resulting pH=-log( 9.358388878 *10^(-5) =4.02787

**pH is 4.03**

Concentration [C2] of second solution

[C2]= 10^(-5.5)=3.16227766*10^(-6) mol/L of [H+]

Total number of moles H+ in the new solution

Concentration of H+ in new solution

[C]=

Resulting pH=-log( 9.358388878 *10^(-5) =4.02787

Jun 15, 2014 | Office Equipment & Supplies

Why not tell us what you are trying to do?.

n= number of moles

m= mass in g.

M= atomic or molar mass (in g/mol)

n=m/M (in mol)

To enter Avogadro's number you use the EE (*10^x) or Exp

6.02 [EE]23 No 10 should appear.

To calculate the atomic/molar mass use the Periodic table.

n= number of moles

m= mass in g.

M= atomic or molar mass (in g/mol)

n=m/M (in mol)

To enter Avogadro's number you use the EE (*10^x) or Exp

6.02 [EE]23 No 10 should appear.

To calculate the atomic/molar mass use the Periodic table.

Oct 30, 2013 | Texas Instruments TI 30XIIS Scientific...

Hydrogen gas is a diatomic substance: Each molecule contains two hydrogen atoms.

A mole of hydrogen gas contains 6.02*10^23 hydrogen molecules, but since each molecule contains 2 hydrogen atom you have**2*6.02*10^23=?**

The formula for nitric acid is HNO_3. It contains 5 atoms regardless of the elements

Now, 0.25 moles contain 0.25* 6.02*10^23 molecules oh nitric acid, ans since each molecule contains 5 atoms then 0.25 moles contain**0.25*5*6.02*10^23=?** atoms

0.25*1*6.02*10^23 of hydrogen atoms, and the same number of nitrogen atoms and 0.25*3*6.02*10^23 oxygen atoms.

A mole of hydrogen gas contains 6.02*10^23 hydrogen molecules, but since each molecule contains 2 hydrogen atom you have

The formula for nitric acid is HNO_3. It contains 5 atoms regardless of the elements

Now, 0.25 moles contain 0.25* 6.02*10^23 molecules oh nitric acid, ans since each molecule contains 5 atoms then 0.25 moles contain

0.25*1*6.02*10^23 of hydrogen atoms, and the same number of nitrogen atoms and 0.25*3*6.02*10^23 oxygen atoms.

Oct 01, 2013 | Cars & Trucks

U have ur tv original remote look for a button like ViewMode,Aspect Ratio,Picture sizes,Picture Zoom.Tries pushes one of these buttons a couple times,u should able to changes ur tv pictures sizes.

Jun 13, 2011 | Philips 42PFL7403D/27 42 in. LCD TV

This problem is the type often encountered in a first semester chemistry course, whether in high school or college.

It is very easy to solve if you already understand the following concepts, and if you also have had lots of practice applying them!

To solve this problem you must be aware of the**balanced chemical equation**, as follows:

Notice that there is**one** C, **one** O2, and **one** CO2.

*(For convenience in my post, I am writing the subscripts on the same line,)*

"One what?" You might ask. It can be__one__ "**molecule**" for each the O2 and CO2; and __one__ "**atom**" of C. For the method of **stoichiometry** in this problem, we should use **one** "mole." So, there is **one** __mole__ of C, **one** __mole__ of O2, and **one** __mole__ of CO2. In other words, the **PROPORTION** of these reactants (C and O2) and product (CO2), is **one to one to one**, also written **one:one:one** (or 1:1:1).

As I am assuming you know, these numbers are called the "**coefficients**" of the balanced chemical equation.

To calculate the**mass in grams of O2** (the unknown *quantity*), we will need only the moles of O2 and the moles of one other substance from the balanced chemical equation. Since we are given the mass of C, we can use the moles of C.

TO WORK A STOICHIOMETRIC problem using a balanced chemical equation, YOU MUST INCLUDE THE MOLES OF TWO SUBSTANCES from the chemical equation in your calcuation, in this case, moles of C and moles of O2. As you know, you can calculate moles of C from the mass (in grams) of C that was given in this problem. You must calculate the moles of O2 by use of the following math expression:

mol O2 = (MOLE RATIO of O2 to C) x mol C.

(Notice that I have used the official abbreviation of moles, which is "mol".)

I will show you how to calculate the MOLE RATIO a moment.

Once you have caculated the moles of O2, you will have only one more calculation to do to calculate the number of grams of O2. Since I am assuming you already know how to calculate grams from moles, I will skip this step until my closing summary.

The MOLE RATIO can be calculated as follows:** The calculated mole ratio will always equal the ratio of corresponding coefficients.**

Notice that the mole ratio actually represents the ratio of the coefficients that are in the balanced chemical equation, and is written in the same order as are the calculated (calc'd) moles.

Therefore,

So, you can see that the number of moles of O2 = calculated moles of C.

The**moles of C** were calculated as follows:

It is very easy to solve if you already understand the following concepts, and if you also have had lots of practice applying them!

- PROPORTIONS
- THE MOLE
- The LAW OF CONSERVATION
- CHEMICAL EQUATIONS
- BALANCED CHEMICAL EQUATIONS
- STOICHIOMETRIC CALCULATIONS
- SIGNIFICANT FIGURES, PRECISION, & ROUNDING OFF CALCULATED QUANTITIES

To solve this problem you must be aware of the

Notice that there is

"One what?" You might ask. It can be

As I am assuming you know, these numbers are called the "

To calculate the

TO WORK A STOICHIOMETRIC problem using a balanced chemical equation, YOU MUST INCLUDE THE MOLES OF TWO SUBSTANCES from the chemical equation in your calcuation, in this case, moles of C and moles of O2. As you know, you can calculate moles of C from the mass (in grams) of C that was given in this problem. You must calculate the moles of O2 by use of the following math expression:

mol O2 = (MOLE RATIO of O2 to C) x mol C.

(Notice that I have used the official abbreviation of moles, which is "mol".)

I will show you how to calculate the MOLE RATIO a moment.

Once you have caculated the moles of O2, you will have only one more calculation to do to calculate the number of grams of O2. Since I am assuming you already know how to calculate grams from moles, I will skip this step until my closing summary.

The MOLE RATIO can be calculated as follows:

Notice that the mole ratio actually represents the ratio of the coefficients that are in the balanced chemical equation, and is written in the same order as are the calculated (calc'd) moles.

Therefore,

So, you can see that the number of moles of O2 = calculated moles of C.

The

The calculated moles of C = [grams C] / [MW of C] = [188 g C / 12.0 g] = 15.7 mol C

Therefore, there are 15.7 moles of O2, as calculated from use of the math expression (with boxes) shown above.

The wanted GRAMS of O2 are calculated from MW of O2 times its number of moles, as follows:

Grams of O2 = 32.0 g O2 x 15.7 mol O2 = **502 grams O2**.

For your convenience as you make conversions among grams and moles, you may use the following memory aid:

To go from moles to grams, just notice that moles are beside MW, so multiplying them will give grams (moles x MW = grams). If you want to go from grams to moles, notice that grams are over MW, so (grams/MW) = moles. If you should want to calculate MW, you would divide grams by moles, (grams/moles) = MW.

One more tip: When doing calculations, use at least the same number of significant figures (*sig figs*) in the molecular weights (and atomic weights) as is in the given numerical data. That is, since there are three sig figs in 188 grams of C, use 12.0 as the atomic weight of C, and 44.0 as the MW of CO2. You would obtain the same answer using 12 and 44, but as you may know, following this rule would be more important in other problems in which the digit past the decimal is not zero. This allows proper rounding off of the final calculated quantity.

**In Summary**

- Write the chemical equation, and balance it.
- Use MOLES to do the necessary stoichiometric calculations. Convert grams to moles to make this possible.
- Multiply the calculated moles of the given amount of chemical compound (or element) by the COEFFICIENT RATIO of the two compounds involved in the problem.
- Convert moles to grams if grams are asked for by the problem.
- Round off correctly.
- Show the answer quantity, which should always show both numerical value AND unit.

Suggestion: Try out this method to other problems involving a chemical reaction. Good luck!

***

Dec 12, 2010 | Scientific Explorer My First Chemistry Kit

The WS-55859 does not have a "Lamp" it is a 2001 rear projection set with three picture tubes.

Have (if this is the correct model) you tried the self check this set has?

Also do you have this problem on all inputs--tuner, and various rear inputs?

The more you tell me the better I can help you.

Got a IM?

SD TECH

Have (if this is the correct model) you tried the self check this set has?

Also do you have this problem on all inputs--tuner, and various rear inputs?

The more you tell me the better I can help you.

Got a IM?

SD TECH

Feb 23, 2010 | Mitsubishi WS-55859 55" Rear Projection...

I have sound on video 2, and when on a static channel but not on the usual channel 3 which is TV. Just stoppen working one day and cant figure it out... Please help!

Oct 19, 2008 | Televison & Video

Oct 22, 2007 | RCA F32450 32" TV

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