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3 years ago

7. Show the numerical setup of titration formula (MA)(VA)=(MB)(VB)

Chemistry

1 answer:

nirvana33 [79]3 years ago

4 0

Answer:

i believe it is 17x=(4)x*4 +11*(0)

Explanation:

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In the vertical columns of the periodic table, as the atomic numbers increase:

kolezko [41]

A.) The number of electron shells increases.

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3 years ago

Read 2 more answers

SELECT THE STATEMENT ABOUT ELECTRONS AND ATONS THAT IS TRUE 1.EACH ORBITS HOLDS A FIXED NUMBER OF ELECTRONS 2.THE N=1 ORBIT CAN

mel-nik [20]

1. <span>EACH ORBITS HOLDS A FIXED NUMBER OF ELECTRONS </span>

7 0

3 years ago

Suppose that you lowered the temperature of a gas from 100˚C to 50 ˚C. By what factor do you change the volume of the gas?

Maru [420]

Answer:- Volume decreases by a factor of 1.15.

Solution:- At constant pressure, volume of the gas is directly proportional to the the kelvin temperature.

The equation is written as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where, $V_1$ is the volume at initial temperature $T_1$ and $V_2$ is the volume at final temperature $T_2$ .

Temperature must be in kelvins. So, let's convert both the temperatures to kelvin.

To convert degree C to kelvin we add 273.

So, $T_1$ = 100 + 273 = 373 K

$T_2$ = 50 + 273 = 323 K

The equation could also be written as:-

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

$\frac{V_1}{V_2}=\frac{373}{323}$

$\frac{V_1}{V_2}$ = 1.15

From here we could say that the volume decreases by a factor of 1.15.

For example if the initial volume $V_1$ is 1 L then final volume $V_2$ will be $\frac{1L}{1.15}$ that is 0.87 L.

4 0

3 years ago

Calculate the density of a liquid if 58.9 ml of it has a mass of 46.08 g. answer in units of g/ml.

alexira [117]

Density of a solution is mass of solution per unit volume
Density = mass/volume
mass of solution is 46.08 g
volume of solution is 58.9 mL
since mass and volume is known, density can be calculated
density = 46.08 g / 58.9 mL = 0.78 g/mL

8 0

3 years ago

Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz

Hoochie [10]

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

5 0

3 years ago