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Savatey [412]
3 years ago
7

What is the percent by mass of potassium in KFe(CN)?

Chemistry
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer:

Explanation:

1. Add the atomic mass of all the elements.

39+55.8+12+14= 120.8

2.Divide atomic mass of potassium by total atomic mass

39/120.8= .323

2. Multiply by 100

.323*100= %32.3 %32.3

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Determine the concentration of an HBr solution if a 45.00 mL aliquot of the solution yields 0.5555 g AgBr when added to a soluti
amid [387]
Molecular weight of AgBr = 187.7
moles of Ag = \frac{0.5555}{187.7} = 2.96 x 10^{-3}
moles of Br = moles of Ag = 2.96 x 10⁻³ mol
concentration of HBr (Molarity) = conc. of Br (strong acid) = \frac{2.96 x 10^{-3} }{45 x 10^{-3} } = 0.0658 mol/l
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3 years ago
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The Law of Conservation of Matter describes how mass is not lost during a chemical reaction.
weeeeeb [17]

False, it states that matter can be nether created nor destroyed

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Define the boiling and evaporation using kinetic theory
expeople1 [14]

Answer:

Explanation:

By the kinetic molecular theory (particle model), all matter consists of particles, there are spaces between the particles, the particles are in constant random motion, and there are forces of attraction and repulsion between the particles.

Furthermore, temperature is defined to be a measure of the average kinetic energy of the particles.

Evaporation is a change of phase from liquid to gas explained as follows :

When particles in the liquid phase are heated, they gain kinetic energy and move faster and further apart. Eventually they have enough energy to escape the forces of attraction holding them together in the liquid phase and they move very fast and far from each other and exist in the gaseous phase.

5 0
2 years ago
If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
Korvikt [17]

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

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
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