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Anastaziya [24]
3 years ago
9

(3.20x10^4)(0.2402) express your answer in appropriate significant figures.

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
7 0
7686.4 is the awnser
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A 6.50x10^-5 m solution of potassium permanganate has a percent transmittance of 27.3% when measured in a 1.15 cm cell at a wave
mihalych1998 [28]
For this problem, we use the Beer Lambert's Law. Its usual equation is:

A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
C is the concentration of the sample solution

As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:

A = 2 - log(%T)
A = 2 - log(27.3)
A = 0.5638
7 0
3 years ago
9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M
antiseptic1488 [7]

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

5 0
3 years ago
A 5 grams sample of liquid water evaporates into gaseous water in a closed container, expanding in the process. What is the mass
padilas [110]

We have get the mass of gaseous water after evaporation in a closed container.

The mass of water vapor after evaporation is 5 grams.

In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.

Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.

Randomness of  molecules increases in gaseous state than liquid state.  


5 0
3 years ago
The solubility of a compound with a molecular weight of 233.2 is determined to be 27.9 g in 125.0 g of water at 398 K. Express t
sp2606 [1]

Answer:

0.960 m

Explanation:

Given data

  • Mass of the solute: 27.9 g
  • Molar mass of the solute: 233.2 g/mol
  • Mass of the solvent: 125.0 g = 0.1250 kg

First, we will calculate the moles of solute.

27.9 g × (1 mol/233.2 g) = 0.120 mol

The molality of the compound is:

m = moles of solute / kilograms of solvent

m = 0.120 mol / 0.1250 kg

m = 0.960 m

5 0
3 years ago
What is always true of a strong acid?
sveta [45]

Strong acids provide more H+ in the aqueous solution than the weak acids. This excess amount of H+ conducts more electricity in the solution.

Explanation:

The strong acid provides more H+ ion whereas the weak acids produce less H+ in the aqueous solution resulting in less amount of electricity conduction in the solution.

Weak acid- lesser amount of H+ produced

Strong acid- greater amount of H+ ( proton) produced.

4 0
3 years ago
Read 2 more answers
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