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2 years ago

15

¿Qué bacterias existen según su nutrición?

1 answer:

Maurinko [17]2 years ago

7 0

Answer:

autotrofas estrictas son aquellas bacterias incapaces de crecer usando materia orgánica como fuente de carbono. Mixotrofas son aquellas bacterias con metabolismo energético litotrofo (obtienen energía de compuestos inorgánicos), pero requieren sustancias orgánicas como nutrientes para su metabolismo biosintético.

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Generally, most water reabsorption in a nephron occurs at the:_________

Allisa [31]

Answer:

proximal convoluted tubule and is called obligatory water reabsorption.

4 0

3 years ago

Lice are parasitic insects that live on the scalp, eyebrows, and eyelashes of humans

earnstyle [38]

Answer:

Most likely A

Explanation:

Its a more plausible answer

5 0

3 years ago

What is the molarity of HCl if 34.81 mL of a solution of HCl contain 0.3297 g of HCl?

Temka [501]

Answer:

0.2598 M

Explanation:

Molarity is mol/L, so we have to convert the grams to moles and the mL to L. To convert between grams and moles you need the molar mass of the compound, which is 36.46g/mol.

$0.3297gHCl*\frac{1molHCl}{36.46gHCL} = 0.00904279 mol HCl$

$34.81 mLHCl * \frac{1L}{1000L} = 0.03481 L HCl$

$\frac{0.00904279molHCl}{0.03481 L HCl} = 0.25977554 mol/L$

Round to the lowest number of significant figures = 0.2598 M

6 0

3 years ago

NEED HELP ASAPPP<br><br> What is the number of beryllium atoms in 78.0g of Be?

polet [3.4K]

Answer: number of atoms is 5.21 · 10^24

Explanation: Atomic mass of Be is 9.012 g/mol.

Number of moles n = m/M = 78.0 g / 9.012 g/mol =

Multiply this with Avogadro number Na = 6.022*10^23 1/mol

4 0

3 years ago

The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

5 0

3 years ago