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3 years ago

7

3.4Why is this cocktail not the perfect medication to treat infections by thesebacteria?

1 answer:

ryzh [129]3 years ago

8 0

Answer:

many bacteria are resistant to penicillin and other β-lactam antibiotics - some bacteria have beta-lactamase enzymes and are able to break down the antibiotics - as a result, many drugs are combined with other drugs that are β-lactamase inhibitors.

Explanation:

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Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what

Drupady [299]

Answer:

<h3>160 cm</h3>

Explanation:

6 0

3 years ago

Which of the following is a problem caused by acid rain?

marissa [1.9K]

Answer:

1

Explanation:

is a factor in global warming

3 0

3 years ago

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

<u>Answer:</u> The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

Technetium-99 is an ideal radioisotope for scanning organs because it has a half-life of 6.0 h and is a pure gamma emitter. supp

Nadusha1986 [10]

Answer: 20 mg Te-99 remains after 12 hours.

Explanation: N(t) = N(0)*(1/2)^(t/t1/2)

N(t) = (80 mg)*(0.5)^(12/6)

N(t) = 20 mg remains after 12 hours

3 0

3 years ago

zepelin [54]

Because the ring is hollow

3 0

3 years ago