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2 years ago

Do you think that it is possible for that viruses can grow and respond to changes in their environment? Why or why not?

Chemistry

2 answers:

lakkis [162]2 years ago

7 0

Answer:

So... yes :)

Explanation:

Viruses are nonliving agents that depend on the host cell that they are infecting to reproduce. They are bundles of genetic material enclosed in a capsid (some viruses have an additional layer called the viral envelop). Viruses can only thrive and replicate inside the environment of a living cell of other organisms. Viruses adapt to the environment (the cell) they are in by infecting the entire cell. Viruses can infect other nearby cells by infecting its genetic code (either DNA or RNA) and spread. Viruses develop and spread by attaching its protein tail to a receptor site on the host cell wall and injecting its genetic material into the host cell. The virus can then take over the cell and spread to infect other cells in a similar way. This is how viruses replicate. Outside of the host cell, most viruses are inert.

n200080 [17]2 years ago

4 0

Yes, I do think it’s possible because if viruses spread, (like corona), then it could definitley evolve into something worse than a sickness.

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Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bon

vampirchik [111]

Answer:

(A) An investor will exercise the option to put the bond if yields rise significantly

Explanation:

A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.

Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.

5 0

3 years ago

Use this balanced equation to help you solve:

valentina_108 [34]

1. 842g of NaOH will form 547.3 g of Al(OH)₃

2. The yield is 93.55%

Explanation:

3NaOH + Al → Al(OH)₃ + 3Na

Molar mass of NaOH = 40 g/mol

Molar mass of Al = 27 g/mol

Molar mass of Al(OH)₃ = 78 g/mol

According to the balanced equation:

3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃

The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1

3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃

120 g of NaOH + 27g of Al → 78 g of Al(OH)₃

120g of NaOH form 78g of Al(OH)₃

1g of NaOH will form $\frac{78}{120}$ g of Al(OH)₃

842g of NaOH will form $\frac{78}{120} X 842 g$ of Al(OH)₃

= 547.3 g of Al(OH)₃

Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃

2. Only 512 g of Al(OH)₃ is formed

Yield % = ?

$Yield = \frac{512}{547.3} X 100\\\\Yield = 93.55$

Therefore, the yield is 93.55%

5 0

3 years ago

What tissue is needed for the lungs organ

Vadim26 [7]

Answer:

The alveoli are responsible for the spongy nature of the lung. These alveoli are lined by flattened epithelial cells called pneumocytes with a single opening. The alveolar wall or septum is made up of three tissue components: surface epithelium, supporting tissue, and an extensive network of continuous capillaries

4 0

2 years ago

What is the standard cell notation for a galvanic cell made with silver and nickel?

MA_775_DIABLO [31]

Answer:

Ni(s)∣∣Ni2+(aq) ∣∣∣∣ Ag+(aq)∣∣Ag(s)

Explanation:

Start by finding the standard reduction potential for the Ag+ and Ni2+ ion. Normally, the values are listed at the back of most chemistry textbooks.

Ag+(aq)+1e−→Ag(s) Eo=0.80 V
Ni2+(aq)+2e−→Ni(s) Eo=−0.23 V

In the galvanic cell, the reaction is spontaneous and for a spontaneous reaction Eocellmust be a positive quantity.

Eocell=EoAnode+Eocathode

Manipulate the two equations so that Eocell is positive. Note that the anode is the site of oxidation (where electrons are lost) and the cathode (where electron are gained) is the site for reduction.

Ag+(aq)+1e−→Ag(s) Eo=0.80 V
Ni(s)→Ni2+(aq)+2e− Eo=0.23 V

2×{Ag+(aq)+1e−→Ag(s)} Eo=0.80 V (Cathode)
Ni(s)→Ni2+(aq)+2e− Eo=0.23 V (Anode)
 −−−
2Ag+(aq)+Ni(s)→Ni2+(aq)+2Ag(s) Eocell=1.03 V

Start with the anode components (site of oxidation) - the cathode components are listed to the right.

The single vertical lines indicate the boundary (phase difference) between solid Niand Ni2+ ions in the aqueous solution of the first compartment and between solid Ag and Ag+ ions present in the aqueous solution of the second compartments.

The double vertical lines refer to the salt bridge - note that the salt bridge must be an inert salt to both ions present in both compartments of the galvanic cell ...

5 0

3 years ago

How many grams of nickel(II) chloride do you need to precipitate 503 mg of silver chloride in the reaction between nickel(II) ch

maxonik [38]

From the information provided in the question, the mass of NiCl2 required is 0.23 g.

NiCl2(aq) + 2AgNO3(aq) -------> 2AgCl(s) + Ni(NO3)2(aq)

Number of moles of AgCl= Mass/molar mass

Molar mass of AgCl= 143 g/mol

Substituting values;

Number of moles = 503 × 10^-3g/143 g/mol= 0.00352 moles

Since 1 mole NiCl2 precipitates 2 moles of AgCl

x moles of NiCl2 precipitates 0.00352 moles of AgCl

x = 1 mole × 0.00352 moles/2 moles

x = 0.00176 moles of NiCl2

Mass of NiCl2= 0.00176 moles × 130 g/mol = 0.23 g

Learn more: brainly.com/question/2510654

5 0

2 years ago