Number 14? Please :)

If an acid has a ka=1.6x10^-10, what is the acidity of the solution?

LUCKY_DIMON [66]

The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).

5 0

3 years ago

Select the statements that correctly describe the term viscosity. a.The viscosity of each substance is a constant value that wil

KatRina [158]

Answer:

D. VISCOSITY can be defined as a substance's ability to resist flow.

Explanation:

This is the statement that correctly describe the term viscosity. Viscosity explains why substances or fluids resist flow. It shows that internal friction or resistance to flow revealed by a substance. A substance with high viscosity or internal friction resist flow and one with low viscosity flow easily. Viscosity of a substance change base on various factors affecting the substance such as pressure, temperature, surface area. Substances like water has low viscosity explaining why it flows easily.

8 0

3 years ago

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .

3 0

3 years ago

A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC

Rudiy27

The mass of nitrogen collected is mathematically given as

M-N2=0.025gram

<h3>What is the mass of nitrogen collected?</h3>

Question Parameters:

A sample weighing 2.000g

the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).

T=26.3c=299.3K

Pressure=745mmHg=745torr

Pressure of N2=745-25.2=719.8torr

Generally, the equation for the ideal gas is mathematically given as

PV=nRT

Therefore

719.8/760=45.6/1000=n*0.0821*299.3

n=0.00176*14

In conclusion, the Mass of N2

M-N2=0.00176*14

M-N2=0.025gram