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PANO POH MAG REPORMAT NG CHEERY MOBILE t18 NG HUHUNG POH EH PANO POH

Dec 07, 2013 | Cell Phones

The pH scale is a logarithmic scale used to measure the concentrations of hydrogen/Hydronium ions in an aqueous solution.

If c[H] is the concentration of hydrogen ions in mol/L, the pH is defined as

pH=-log(c[H]).

Inversely, the concentration is given by

c[H]=10^(-pH)

The pOH=14-pH

If c[H] is the concentration of hydrogen ions in mol/L, the pH is defined as

pH=-log(c[H]).

Inversely, the concentration is given by

c[H]=10^(-pH)

The pOH=14-pH

Jan 19, 2012 | Office Equipment & Supplies

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Thank you.

Thank you.

Sep 13, 2011 | Computers & Internet

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May 18, 2011 | Computers & Internet

sorry don't understand language

Dec 01, 2010 | Microsoft Windows XP Professional

I am going to assume your term, "nf," which I do not recognize as an acceptable unit of the metric system, is meant to be **molarity** **(M)**. So, it seems to me that you are asking for the __molarity__ of the solution that is indicated in the problem.

To calculate the molarity of the solution of NaOH in this problem, one must apply the ratio, moles per liter of solution, the definition of molarity (moles/L).

A 27.5% solution means (27.5 grams) / (100 grams of solution), which in this case is equivalent to

27.5 g / 100 mL solution, because the density of water is 1.00 g/mL.

Since 100 g solution is specified, that means there are 27.5 g of NaOH in the solution (100 mL volume, or 0.100 L).

Moles of NaOH in solution = [27.5 g NaOH / 40.0 g mol NaOH/ mol NaOH] = 0.688 mol NaOH.

__Therefore, molarity of NaOH = mol NaOH / L = (0.688 mol NaOH / 0.100 L)__ = **6.88 M NaOH**.

###

To calculate the molarity of the solution of NaOH in this problem, one must apply the ratio, moles per liter of solution, the definition of molarity (moles/L).

A 27.5% solution means (27.5 grams) / (100 grams of solution), which in this case is equivalent to

27.5 g / 100 mL solution, because the density of water is 1.00 g/mL.

Since 100 g solution is specified, that means there are 27.5 g of NaOH in the solution (100 mL volume, or 0.100 L).

Moles of NaOH in solution = [27.5 g NaOH / 40.0 g mol NaOH/ mol NaOH] = 0.688 mol NaOH.

###

Jun 02, 2010 | Scientific Explorer My First Chemistry Kit

First you have pH+pOH=14

Thus pOH=14-pH.

The problem with you question is that you do not specify the meaning of the 1.2 x 10^(-3). Is it the molar concentration of the hydrpbromic acid HBr.

I am going to answer your question but I must make some assumptions. You can imitate the reasonning to obtain your answer, once you have decided or looked up what the numbers mean.

Lets assume that HBr concentration is 1.2x10^(-3) mol/L. Since it is a strong acid, it dissociates completely, releasing 1.2x10^(-3) mol/L H+/H3O+ ion.

Thus the pH is

pH=-log(1.2x10^(-3)).

The result is 2.92

Here is a screen capture of the actual calculation. The minus signs are the change sign (-) to the right of the dot. To enter the power of 10 (represented by the E in display) use the (2nd)(EE) key sequence. To access the content of (Ans) memory, press (2nd) (change sign)

Thus pOH=14-pH.

The problem with you question is that you do not specify the meaning of the 1.2 x 10^(-3). Is it the molar concentration of the hydrpbromic acid HBr.

I am going to answer your question but I must make some assumptions. You can imitate the reasonning to obtain your answer, once you have decided or looked up what the numbers mean.

Lets assume that HBr concentration is 1.2x10^(-3) mol/L. Since it is a strong acid, it dissociates completely, releasing 1.2x10^(-3) mol/L H+/H3O+ ion.

Thus the pH is

pH=-log(1.2x10^(-3)).

The result is 2.92

Here is a screen capture of the actual calculation. The minus signs are the change sign (-) to the right of the dot. To enter the power of 10 (represented by the E in display) use the (2nd)(EE) key sequence. To access the content of (Ans) memory, press (2nd) (change sign)

Mar 07, 2010 | Texas Instruments TI-83 Plus Calculator

Hello,

You should use the equivalence

**y=10^(x) is equivalent to x=log(y)**

You should not use word that have no meaning. x above is sometimes called antilog, not reverse log.

The relation linkin pH and pOH is as follows

** pH+pOH=14,** thus **pH=14-pOH**

Let c[H+] be the concentation (or if you use the hydronium ion, instead of H+ ) c[H3O+] the concentration of hydronium ion)

c[H+]=c[H3O+]=10^(-pH)

pOH= 6.95 thus pH=14-6.95 = 7.05

If pH=7.05 the c[H+]= 10 ^(-7.05) =8.9125E(-8)

pH=**10^ [X to the power] [(-)]7.05 **

The key/buttom to raise to the power is the one to the right of x squared

Hope it helps

You should use the equivalence

You should not use word that have no meaning. x above is sometimes called antilog, not reverse log.

The relation linkin pH and pOH is as follows

Let c[H+] be the concentation (or if you use the hydronium ion, instead of H+ ) c[H3O+] the concentration of hydronium ion)

c[H+]=c[H3O+]=10^(-pH)

pOH= 6.95 thus pH=14-6.95 = 7.05

If pH=7.05 the c[H+]= 10 ^(-7.05) =8.9125E(-8)

pH=

The key/buttom to raise to the power is the one to the right of x squared

Hope it helps

Sep 10, 2009 | Texas Instruments Office Equipment &...

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