Which of the following statements is not true for Coulomb's law?

A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of

blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of $H^+$ ion = $Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol$

Moles of $OH^-$ ion = $Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol$

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

0.01 mol of $H^+$ ion will react with = $\frac{1}{1}\times 0.01mole$ of $OH^-$ ion

Thus (0.012-0.01)= 0.002 moles of $OH^-$ are left in 20 ml or 0.02 L of solution.

$[OH^-]=\frac{0.002}{0.02L}=0.1M$

$pOH=-log[OH^-]$

$pOH=-log[0.1]=1$

$pH+pOH=14$

$pH=14-1=13$

Thus the pH of resulting solution will be 13

7 0

2 years ago

An unsaturated solution has,

olganol [36]

Answer:

D... I think? Because it is unsaturated so there is more solute than solvent...

8 0

2 years ago

Read 2 more answers

if there are 4 grams of reactant,how many grams of product are produced by the chemical reaction. a.0 grams b.2 grams c.4 grams

Murrr4er [49]

4 grams , mass cannot be created or destroyed. it has to be the same on both sides of the equation

5 0

2 years ago

Find the mass in 2.6 mole of lithium bromide

Tresset [83]

Answer:

2.6 mole of lithium bromide have 225.81 g.

Explanation:

Given data:

Number of moles of lithium bromide = 2.6 mol

Mass of lithium bromide = ?

Solution:

Formula:

<em>Number of moles = mass/ molar mass</em>

Molar mass of lithium bromide = 86.85 g/mol

Mass = number of moles × molar mass

Mass = 2.6 mol ×86.85 g/mol

Mass = 225.81 g

Thus 2.6 mole of lithium bromide have 225.81 g.

5 0

2 years ago

What fraction of the gas phase molecules initially present for 1.0 atm in the chamber are present at 1.0×10−10 torr?

vivado [14]

The fraction of gas phase molecules is calculated by the division of final pressure to the initial pressure.

Fraction = $\frac{P_{final}}{P_{initial}}$ (1)

Here, initial pressure = 1.0 atm

final pressure = $1.0\times 10^{-10} torr$

First, convert torr into atm

1 atm = 760 torr

final pressure = $\frac{1.0\times 10^{-10}}{760}$

= $1.316\times 10^{-13}atm$

Now, put the value of initial and final pressure in formula (1)

Fraction = $\frac{1.316\times 10^{-13}atm}{1.0 atm}$

= $1.316\times 10^{-13}$

Thus, fraction of the gas phase molecules = $1.316\times 10^{-13}$ $\simeq 1.3\times 10^{-13}$

7 0

3 years ago