What are the terms used to describe a location on a sanddun?

| A solution containing 4.48 ppm KMnO4 exhibits

Artist 52 [7]

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = $\frac{4.48 X10^{-3} }{158.03X 1(lit)}$ = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T) ( T = % transmittance)

= - log(0.859)

= 0.06

According to Lambert Beer's law

ε = $\frac{A}{C X l}$

or, ε = $\frac{0.06}{2.83 X 10^{-5}X1 cm }$

or, ε = 2120.14 cm⁻¹M⁻¹

Where

ε = Molar absorptivity

A = absorbance

C = Molar concentration of KMnO₄ solution

l = length

6 0

3 years ago

Complete and balance the following acid-base equations:

Alchen [17]

Answer:

a) $HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

b) $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

c) $Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

Explanation:

a)

when $HClO_4$ is added to LiOH, lithium chlorate and water is formed.

$HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous $H_2SO_4$ is added to NaOH, $Na_2SO_4$ and $H_2O$ is formed. It is a neutrilization reaction.

$H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

Now, it can been seen that all the atoms are balanced.

c) When $Ba(OH)2$ reacts with HF, baroum fluoride and water is formed.

$Ba(OH)2 + HF\rightarrow BaF_2 + H_2O$

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

$Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O$

In order to balance H and O on either side, multiply H2O by 2.

$Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

7 0

3 years ago

The molar mass of glucose is 180.2 g/mol. How many grams of glucose will be produced when 132.0 g of CO2 reacts with an excess o

andriy [413]

Answer:

The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams

Explanation:

The chemical equation for the reaction is

6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂

From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas

The molar mass of CO₂ = 44.01 g/mol

There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles

However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆

and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of

mass of 1 mole C₆H₁₂O₆ = 180.2 g

mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams

Mass of glucose produced = 90.1 grams

7 0

3 years ago

how does the percentage by mass of the solute describe the concentration of an aqueous solution os potassium sulfate

kirza4 [7]

Answer:

Explanation:

In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:

mass percentage = (mass of solute/total mass of solution) x 100%

For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.

3 0

2 years ago

What is the oh−concentration and ph in each solution? (a) 0.225 m koh, (b) 0.0015 m sr(oh)2?

shepuryov [24]

Q1)
We have been given the OH⁻ concentration, therefore we first need to find the pOH value and then the pH value.
pOH = -log [OH⁻]
pOH = -log (0.225 M)
pOH = 0.65
pH + pOH = 14
pH = 14 - 0.65 = 13.35

Q2)
pOH = -log[OH⁻]
pOH = -log (0.0015 M)
pOH = 2.82
pH + pOH = 14
pH = 14 - 2.82
pH = 11.18

7 0

3 years ago