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3 years ago

What is the percent by mass of potassium in K3Fe(CN)6?

Chemistry

2 answers:

zhenek [66]3 years ago

5 0

Answer:

The percentage of potassium in the given complex is 35.54 %.

Explanation:

Mass of potassium in $K_3Fe(CN)_6$ = 3 × 39.10 g mol=117.3 g/mol

Molar mass of $K_3Fe(CN)_6$ =329.15 g/mol

Percentage of potassium (K) in the the complex:

$\% K=\frac{\text{mass of potassium}}{\text{molar mass of complex}}\times 100$

$\%K=\frac{117.3 g/mol}{329.15 g/mol}\times 100=35.54\%$

The percentage of potassium in the given complex is 35.54 %.

yawa3891 [41]3 years ago

3 0

The percent by mass of potassium in K3Fe(CN)6 is 35.62%.

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elements on the right side of the periodic table differ from the elements on the left side in that elements on the right side

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Element on the right side of the periodic table differ from the elements on the left side in that elements on the right side are non metallic and tends to be gases at room temperature.

Explanation

In the periodic table there element in the right side , left side and those which are in between.

Example of element in the right side is fluorine chlorine, neon, Argon among others.

This element have higher effective nuclear charges and stabilize electrons more effectively.
there electrostatic intermolecular forces are generally weak therefore they exist in liquid or gaseous state.

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3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

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3 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

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2 years ago

Which type of wave do the particles in a medium move parallel to the direction that the waves move

Talja [164]

Longitudinal waves. In a longitudinal wave the particles in the medium move parallel to the direction waves go. A good example can be the p-waves in an earthquake.

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3 years ago

The price elasticity of demand measures a. buyers’ responsiveness to a change in the price of a good. b. the extent to which dem

-BARSIC- [3]

Option A

The price elasticity of demand measures buyers’ responsiveness to a change in the price of a good.

Explanation:

Price elasticity of demand holds the responsiveness of need subsequent a variation in a product's cost. In different terms, it’s a process to comprehend out the responsiveness of buyers to inconstancies in cost. Price elasticity estimates the responsiveness of the measure necessitated or outfitted of a good to a shift in its demand.

The price elasticity of demand is the rate fluctuation in the amount demanded of a good or assistance distributed by the percentage shift in the price. Considering the quantity demanded habitually declines with value, the price elasticity coefficient is essentially forever negative.

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2 years ago