All categories

3 years ago

How do bacteria demonstrate adaption overtime?

1 answer:

5 0

I cant really help you but here is a few articles that may help you.
http://www.icr.org/article/do-bacteria-evolve-resistance-antibiotics/
https://www.ncbi.nlm.nih.gov/books/NBK45713/
http://www.windows2universe.org/cool_stuff/tour_evolution_9.html
http://evolution.berkeley.edu/evolibrary/news/080401_mrsa
http://evolution.berkeley.edu/evolibrary/article/_0/evoscales_03

You might be interested in

The following reaction occurs in a car’s catalytic converter.

blagie [28]

Answer:The correct answer is ;

The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.

Explanation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.

In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.

In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in $N_2$ molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent

In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in $CO_2$ molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.

8 0

3 years ago

Read 2 more answers

A student inflates a balloon with helium then places it in the freezer. The student should expect

g100num [7]

Answer:

The frozen balloon shrank because the average kinetic energy of the gas molecules in a balloon decreases when the temperature decreases. This makes the molecules move more slowly and have less frequent and weaker collisions with the inside wall of the balloon, which causes the balloon to shrink a little.

Explanation:{ BOOM}***

6 0

3 years ago

Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

weeeeeb [17]

Answer:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

Thus, the rate is given as:

$rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}$

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

4 0

4 years ago

A gas has a volume of 450. mL at 55.0 °C. If the volume changes to 502 ml, what is the new temperature?

salantis [7]

Answer:

92.9 °C

Explanation:

Step 1: Given data

Initial volume (V₁): 450. mL

Initial temperature (T₁): 55.0 °C

Final volume (V₂): 502 mL

Step 2: Convert 55.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 55.0 + 273.15 = 328.2 K

Step 3: Calculate the final temperature of the gas

If we assume constant pressure and ideal behavior, we can calculate the final temperature of the gas using Charles' law.

T₁/V₁ = T₂/V₂

T₂ = T₁ × V₂/V₁

T₂ = 328.2 K × 502 mL/450. mL = 366 K = 92.9 °C

7 0

3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

7 0

3 years ago