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Dvinal [7]
3 years ago
8

Which of the following is a unit of volume of solids?

Chemistry
1 answer:
Anika [276]3 years ago
7 0

Answer:

<h3>,,,,,,Cubic meters ,,,</h3><h3 />
You might be interested in
What is the mass sample of 0.0500 moles of zinc chloride ?
vlabodo [156]

Answer:

6.82g

0.59moles

Explanation:

1. What is the mass sample of 0.0500 moles of zinc chloride ?

Given parameters:

Number of moles ZnCl₂ = 0.05moles

Unknown:

Mass of the sample  =  ?

Solution:

To find the mass of a substance using the number of moles, it would be pertinent to understand what mole is.

A mole is a substance that contains the avogadro's number of particles.

It relates to the mass using the expression below;

                Mass of a substance  = number of moles x molar mass

Molar mass of  ZnCl₂;

        Atomic mass of Zn  = 65.4g/mol

                                   Cl = 35.5g/mol

Molar mass = 65.4 + 2(35.5)  = 136.4g/mole

Mass of a substance  = 0.05 x 136.4  = 6.82g

2. How many moles of potassium sulfide are in a 65.50g sample?

Given parameters:

Mass of K₂S  = 65.5g

Unknown:

Number of moles  = ?

Solution:

The number of moles of any substance is related to mass using the expression below;

              Number of moles  = \frac{mass}{molar mass}

Molar mass of K₂S  = 2(39) + 32  = 110g/mol

              Number of moles  = \frac{65.5}{110}   = 0.59moles

8 0
3 years ago
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

3 0
3 years ago
Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single
Alex787 [66]
A I believe would be the answer
3 0
3 years ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
Element, Compound or Molecule<br> Identify each of the images as an element, molecule or compound.
Olin [163]

Answer:

The answer is:

A) Element

B) Molecule

C) Compound

5 0
2 years ago
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