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3 years ago

Which river was so badly polluted that it caught fire several times ?

Chemistry

2 answers:

jekas [21]3 years ago

7 0

Cuyahoga is the answer

tia_tia [17]3 years ago

5 0

The Cuyahoga river has caught fire several times. It was so bad in 1969 that it helps spur an environmental movement.

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Suppose 0.09886 M KOH is titrated into 15.00 mL H2SO4 of unknown concentration until the equivalence point is reached. It takes

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Answer:0.00028917

Explanation:

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3 years ago

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Directions: Show all of your work for each of the following problems.

AysviL [449]

Answer:

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3 years ago

3. Which subatomic particles account for most of an atom's mass?

melamori03 [73]

<h3><u>Answer:-</u></h3>

3. <u>Proton</u> is the subatomic particle which accounts for most of an atom's mass.

4. <u>Electrons</u> are the subatomic particle which accounts for most of atoms volume.

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3 years ago

solution contains 174 grams of lithium bromide in 360 grams of water, what is the mole fraction of lithium bromide

Luba_88 [7]

Answer:

$\large\boxed{\large\boxed{0.091}}$

Explanation:

The<em> mole fraction</em> of a component in a mixture is the number of moles of the component divided by the total number of moles of all the components, multiplied by 100.

The symbol X is used to represent the mole fraction.

$X_A=\dfrac{\text{number of moles of component A}}{\text{number of moles of all the components}}\times 100$

<u />

<u>1. Convert masses in grams to number of moles</u>

Formula:

number of moles = mass in grams / molar mass

a) Lithium bromide

Chemical Formula: LiBr
Molar mass: 86.845 g/mol

$\text{Number of moles}=\dfrac{174g}{86.845g/mol}=2.00mol$

b) Water

Chemical formula: H₂O

Molar mass: 18.015 g/mol

$\text{Number of moles}=\dfrac{360g}{18.015g/mol}=19.98mol\approx 20.0mol$

c) Total number of moles = 2.00 mol + 20.0 mol = 22.0 mol

<u>2. Mole fraction of lithium bromide</u>

$\text{Mole fraction of lithium bromide}=\dfrac{2.00}{22.0}=0.091$

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3 years ago

(a) Given that the path length of the cuvette is 1 cm, what is the extinction coefficient of the 0.020 mM Yellow 5 dye at lambda

garik1379 [7]

Answer:

(a). Kindly check the explanation section.

(b). 24.5 M^-1 cm^-

(c). 0.0318 M

Explanation:

So, we are given the following data parameters which is going to aid us in solving the above Question;

The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.

Using the formula below; we can determine the extinction coefficient as;

Absorbance = extinction coefficient × concentration × path length. -----------(1).

(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.

Then, extinction coefficient = A/ 0.02 × 1 cm.

(b). Making use of the formula above, the extinction coefficient can be Determine as follows;

extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.

(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;

Concentration = absorbance/ extinction coefficient × path length.

Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.

7 0

3 years ago