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Samsung galazy n7000 mic and ringer not working


Posted by Anonymous on


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SOURCE: I have freestanding Series 8 dishwasher. Lately during the filling cycle water hammer is occurring. How can this be resolved

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Posted on Jan 02, 2017

mobil master
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SOURCE: galaxy note n7000 ringer not working

MY dear Your Phone Ringer Default So Please change Ringer
And Ringing Sound Problem will solved

Posted on Aug 31, 2012

  • 13 Answers


So what you are asking for is complete help with on how to use your cell phone, right? If so do one of the following: 1.Read your owner's manual. Or 2. Go to this website( to learn how to use your phone via flash tutorials and you can also download a manual get product support. Their is also a faqs page for the most commonly asked questions as well a page for software available to use on or with your phone. I hope this helps. Good Luck!

Posted on Aug 07, 2007

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SOURCE: samsung Galazy Mini

HI Harry Clark I will tell you the best place to get any user manual is on the manufacture's site. Go to and put in your phone model. Download it to your PC or you can print it out . Tip about printing out manuals is that you can down load the pdf file an save it on your PC. This is a big time saver. Your other choice is go on the phone and type in the make and model and it will come up with a lot of sites you can go to for the manual. Some of the galaxy phones have the newer software installed on it and it will be different from what you are used.

Posted on Nov 10, 2011

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For a college chemistry experiment, students need to prepare a solution containing ethylene glycol. They are each given 12 fluid ounces of a solution containing 26% ethylene glycol, to which they will add...

Here's the math -
ounces Total amount 26% solution = 12 Total amount 22% solution = X Amount of Ethylene in 26% solution = .26 x 12 = 3.12 Amount of Ethylene in 22% solution = .22 x X = .22X Calculating of 24% solution - .24 = (3.12 + .22X) / (12 + X) solving for X - .24X + .24 x 12 = 3.12 + .22X .02X = 3.12 - .24 x 12 .02X = .24 X = 12 But the solution should be obvious because adding like amounts of each produces the solution percentage that falls in the middle.

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1 Answer

What volume of a 2.50m solution of hydrochrolic acid is required to prepare 2.0 litres of a 0.30m solution?

I need to be sure what you meant by the m symbol: M or m. The first is used for molarity and the latter for molality. If it is molality we do need to know the masses per unit volume for the solutions (the so-called densities)

However if you meant 2.5 M and 0.3 M, then the problem is readily solved. I 'll take a step by step method.
2.50 M means that the original solution contains 2.5 mol of solute per 1 L of solution. Let x be the volume of the solution.
The amount of solute ( expressed as a number of moles) isAmount of solute in original solution (in mol)=2.5 (mol/L)*x (L)
Now consider the final solution.
The amount of solute in the final solution is given by:Amount of solute (in mol) =0.3 mol/L*2 L=0.6 mol
However, this last value is exactly the amount of solute the original solution, because you are just putting it in a bigger volume (dilution)
Equating the two amounts of solute we get
2.5 x=0.6 (in mol) and
x=0.6/2,5=0.24 L=240 mL
When concentrations are given in molar units, the foregoing reasoning can be summarized as
initial concentration (c_i) * initial volume (V_i)=final concentration (c_f)*final volume (V_f) or more simply
Using the formula to find the initial volume
V_i=(c_f*V_f)/c_i=(0.3 *2)/2.5=0.24
The starting volume is 0.24 L or 240 mL

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