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insens350 [35]
3 years ago
14

4.8 tons of pig ovaries were required to extract 12.0 mg of estrogen

Chemistry
1 answer:
bearhunter [10]3 years ago
7 0
Is this a question? if so please elaborate.
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Chemical energy.
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3 years ago
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The proportion of dissolved substances in seawater is usually expressed in
LiRa [457]

Answer:

The proportion of dissolved substances in seawater is usually expressed in ppm, ppb or ppt

Explanation:

The concentration of very diluted solutions should be expressed in parts per million, billion or trillion.

ppm = mass from the solute . 10⁶ / mass or volume of the solution

ppb = mass from the solute . 10⁹ / mass or volume of the solution

ppt = mass from the solute . 10¹² / mass or volume of the solution

ppm = mg/kg, μg/g, μg/mL → These are the units

ppb = ng/g

ppt = pg/g

3 0
3 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
When an acid or vase is added to to water,
velikii [3]

Answer:

yes

Explanation:

3 0
3 years ago
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
4 years ago
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