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Naily [24]
3 years ago
14

Which is better?Sexual Reproduction or Asexual Reproduction

Chemistry
1 answer:
Kazeer [188]3 years ago
6 0
I think asexual reproduction is better hope it help
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How emotions affects the quality of paints
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hat is the pressure of CO(g) in equilibrium with the CO2(g) and O2(g) in the atmosphere at 25 ????C? The partial pressure of O2(
Lubov Fominskaja [6]

Answer:

The partial pressure of CO is 5.54x10⁻⁴⁹atm. You shouldn't worry because it is very low pressure

Explanation:

First, the balanced reaction is:

CO + 1/2O₂ → CO₂

The energies of formation are:

ΔG(CO)=-137.168kJ/mol

ΔG(O₂)=0

ΔG(CO₂)=-394.359kJ/mol

The energy of the reaction is:

delta-G_{reaction} =delta-G_{CO_{2} } -(delta-G_{CO} +1/2delta-G_{O_{2} } )\\delta-G_{reaction}=-394.359-(-137.168+0)=-257.191kJ/mol

The expression for calculate the partial pressure of CO is:

p_{CO} =\frac{p_{CO2} }{p_{O_{2} }^{1/2}*exp^{-\frac{delta-G}{RT} }   } \\p_{CO}=\frac{3x10^{-4} }{0.2^{1/2}*exp(-\frac{-257.191*1000}{8.314*298} ) } \\p_{CO}=5.54x10^{-49} atm

5 0
3 years ago
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas, given that the molar mass of
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The molar mass is 16.0 grams 
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Describe how the introduction of a new Predator can cause a species
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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
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