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Nadya [2.5K]
3 years ago
8

¿Que es bacteria y un virus ?es para hoy por favor ayúdeme

Chemistry
2 answers:
swat323 years ago
4 0

i) Las bacterias son organismos microscópicos unicelulares que prosperan en diversos entornos. Estos organismos pueden vivir en el suelo, el océano y dentro del intestino humano.

ii) Un virus es un agente infeccioso submicroscópico que se replica solo dentro de las células vivas de un organismo. Los virus infectan todas las formas de vida, desde animales y plantas hasta microorganismos, incluidas bacterias y arqueas.

VLD [36.1K]3 years ago
3 0
Los antibióticos solo son necesarios para tratar ciertas infecciones causadas por bacterias. Las enfermedades virales no pueden tratarse con antibióticos. Cuando tomar antibióticos no sea lo indicado, pídale a su profesional de atención médica consejos sobre cómo aliviar los síntomas y sentirse mejor.
Enfermedad común
Causa común
¿Se necesitan antibióticos?
Bacteria
Bacteria o virus
Virus
Infección estreptocócica de la garganta (strep)
Sí
Tosferina
Sí
Infección de las vías urinarias
Sí
Sinusitis
Tal vez
Infección del oído medio
Tal vez
Bronquitis/resfriado con congestión de pecho (en niños y adultos)*
No*
Resfriado común/moqueo
No
Dolor de garganta (excepto por infección estreptocócica)
No
Influenza (gripe)
No
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When pineapple juice disolve in water,it forms like what kind of mixtures.​
Digiron [165]

Explanation:

Pineapple juice often shows an unstable cloud and produces a solid precipitate that is not very attractive for consumers. Cloud stabilization by pectin addition is permitted by EU and Codex standards to counteract this effect. This additive must be labeled and its content should not exceed fixed maximum standards (Website of AIJN Code of Practice). Determination of water-soluble pectins by IFU method 26 (Website of International Fruit and Vegetable Juice Association) can be used for control of this parameter. Pectin addition to pineapple juice or juice concentrate, etc. may also be detected after its isolation by 13C isotopic analysis (Hammond, 2006) as explained later.

5 0
2 years ago
Read 2 more answers
It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principall
Aleks [24]

Answer:

Hydrogen: -141 kJ/g

Methane: -55kJ/g

The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.

Explanation:

According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qc + Qb = 0

Qc = -Qb  [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in the temperature

<h3>Hydrogen</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ

The heat released per gram of hydrogen is:

\frac{-162kJ}{1.15g} =-141 kJ/g

<h3>Methane</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ

The heat released per gram of methane is:

\frac{-82kJ}{1.50g} =-55kJ/g

3 0
3 years ago
Which of the following is not an example of a NATURAL polymer?
kondaur [170]
Sugar is not a polymer
5 0
3 years ago
I dont understand this question can u guys help
diamong [38]

Answer: B

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Explanation:

6 0
2 years ago
Read 2 more answers
The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K
BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s):  Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + \int\limits^{T2}_{T1} {DCp(T)} \, dT

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f  for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0
3 years ago
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