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3 years ago

Describe some of the solutions that engineers might propose for maximizing the percent yield of hydrogen from discarded wood.

Chemistry

1 answer:

belka [17]3 years ago

8 0

Tamam öğretmenim teşekkür ederiz çok güzel olmuş mu bir

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39. Which bond is least polar? A) As-C1 B) Bi-CI C) P-CI D) N-C1

garik1379 [7]

Answer:

D) N-Cl.

Explanation:

The electronegativity of the elements in Group 5 (N. P As and Bi) decreases as we go down the Group so N is the most electronegative and chlorine is less able to draw electrons away from N than from the other elements in Group 5.

3 0

3 years ago

describe characteristics of different climate zones as they relate to latitude,elevation,and proximity to bodies of water. (scie

g100num [7]

They or on the southern hysteric watch it on youtube (latitude and longitude song 1 direction remix by the history teachers)

7 0

3 years ago

It the mass of a material is 46 grams and the volume of the material is 8 cm ^3 What would the density of the material to be

Nat2105 [25]

5.75 Grams per cm^3

You do mass divided by volume

3 0

3 years ago

The absorbance of an equilibrium mixture containing FeSCN2 was measured at 447 nm and found to be 0.347. What is the equilibrium

Ksenya-84 [330]

Answer:

The concentration is $C = 1.11 mol/L$

Explanation:

From the question we are told that

The absorbance is $A = 0.347$

The length is $l = 447 nm = 447 *10^{-7} \ cm$

Generally absorbance is mathematically represented as

$A = \epsilon* C * l$

where $\epsilon$ is the molar absorptivity of FeSCN2 with a value $\epsilon = 7.0*10^3 L/cm/mol$

and $C$ is the equilibrium concentration of FeSCN2

$C = \frac{A}{\epsilon * l }$

substituting values

$C = \frac{0.347}{7.0*10^{3} * 447 *10^{-7} }$

$C = 1.11 mol/L$

5 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

3 years ago