What is the pH of a 6.4*10-10 m oh- solution?

A 50.0 50.0 mL solution of 0.127 0.127 M KOH KOH is titrated with 0.254 0.254 M HCl HCl. Calculate the pH of the solution after

Brilliant_brown [7]

Answer:

Explanation:

THE CORECT QUESTION

A 50.0 mL solution of 0.127 M KOH is titrated with 0.254 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.

SOLUTION

Get the concentration of the HCl first using titration formula

CA X V A / CB VB = NA/ NB

Equation of reation; KOH + HCl => KCl + H2O

CA = 0.254 M

CB = 0.127

VA = 1/0.254 = 3.937

CA (after the addition) = 0.127 x 50 / 3.937

= 1.612 M

But pH = - Log[hydrogen ion]

= -log 1.612

=

8 0

2 years ago

At sea level water usually boils at 100. °C and 760. mmHg pressure. On Mt Whitney, the pressure is about 560. mmHg. At what temp

WITCHER [35]

Answer:

91.7°C

Explanation:

We suppose you have a formula to work from. However, that is not supplied with this problem statement, so we looked one up.

The formula in the attachment is supposed to have good accuracy in the temperature range of interest. It gives vapor pressure of water in kPa, not mmHg, so we needed the conversion for that, too.

560 mmHg corresponds to about 74.66 kPa. The attached "Buck equation" formula is used to find the corresponding temperature. The exponential equation could be solved algebraically using logarithms and the quadratic formula, but we choose to find the solution graphically.

Water boils at about 91.7 °C on Mt. Whitney.

7 0

2 years ago

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 1.731.

Amanda [17]

Answer:

The answer to your question is letter c) 6.09 g of sodium and 9.38 g of chlorine.

Explanation:

This problem is solve using rule of three

We know that the proportion Sodium to Chloride is 1 to 1 in sodium chloride, so we have to look for this proportion in the options

AM Sodium = 23 g

AM Chlorine = 35.5 g

Sodium Chlorine

23 g ---------------- 1 mol 35.5 g -------------- 1 mol

1713.73g ------------- x 2666.6 g ------------- x

x = 1713.73/23 = 74.51 x = 2666.6/35.5 = 75.12

These values are very similar, we have to look for the proportion in the options

a) 6.09g of sodium = 0.26 mol

4,87 g of chlorine = 0.14 mol These numbers are not very similar

b) We have 0.26 mol of Na

0.037 mol of Cl This is not the answer

c) We have 0.26 mol of Na

0.26 mol of Cl These numbers are the same, the proportion is 1:1, this is the answer

d) We have 0.26 mol of Na

0.36 mol of Cl This is not the answer

8 0

3 years ago

Relative dating determines the relative age of fossils according to

KiRa [710]

The answer Is C.the layer of rock they're found in

7 0

2 years ago

Read 2 more answers

It's 32°F (0°C) outside under normal atmospheric conditions (1 atm) at a stunt

Korolek [52]

The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass ( $M$ ), in kilograms per kilomole, and mass of the gas ( $m$ ), in kilograms. The volume occupied by the gas ( $V$ ), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

$V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}$ (1)

Where:

$R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
$T$ - Temperature, in Kelvin
$P$ - Pressure, in kilopascals

If we know that $m = 0.239\,kg$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $T = 273.15\,K$ , $P = 101.325\,kPa$ and $M = 2.02\,\frac{kg}{kmol}$ , then the volume of the balloon is:

$V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}$

$V = 2.652\,m^{3}$ ( $2652\,L$ )

The volume of the balloon is approximately 2652 liters. $\blacksquare$

To learn more on ideal gases, we kindly invite to check this verified question: brainly.com/question/8711877

4 0

2 years ago