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3 years ago

Chemical substances that dissolve in water or react with water to release ions are known as

1 answer:

8 0

The answer is electrolytes. Strong electrolytes like strong acids, strong bases and salts dissociate completely into ions when dissolved and no neutral molecules are present in their solution. Weak electrolytes like weak acids and weak bases do not completely ionize in water and some neutral molecules are present in their solution, while nonelectrolytes do not dissociate into ions when in solution at all.

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What distinguishes science from other disciplines that do not use scientific methods?

rewona [7]

Correct Answer: A. Science relies on observation, measurement, and experimentation.

4 0

2 years ago

An FAS calibration curve (absorbance, y-axis vs. FAS in g/mL) gave an equation he regression line of: y-3678(x)+0.056. If an unk

mezya [45]

Answer:

FAS concentration = 1.61*10^-4M

Explanation:

Beer Lambert's law relates the absorbance (A) of a substance to its concentration (c) as:

$A = \epsilon lc----(1)$

where ε = molar absorption coefficient

l = path length

A plot of 'A' vs 'c' gives a straight line with slope = εl

In addition absorbance (A) is related to % Transmittance (%T) as:

A = 2-log%T----(2)

For the FAS solution, the corresponding calibration fit is given as:

y = 3678(x) + 0.056

This implies that the slope = εl = 3678

It is given that %T = 25.6%

$A = 2-log(25.6)=0.592$

Based on equation(1):

$c = \frac{A}{\epsilon l}=\frac{0.592}{3678}=1.61*10^{-4}M$

8 0

2 years ago

Physical Science Structures and Properties of Matter Portfolio

daser333 [38]

Answer:

true

Explanation:

8 0

3 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo

RSB [31]

Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹) because it is already a common factor to all.

The LCM is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

4 0

3 years ago