Complete the table by classifying each property as either a physical or chemical property

suter [353]

If you notice in the graph for antibiotic A, the number of bacteria actually INCREASES as time increases after the antibiotic was given. In the second graph, the amount of bacteria increases just a little bit (likely as the antibiotic sets in) and then decreases until no bacteria is left at all. This means that the antibiotic was the most successful because not only did the amount of bacteria decrease over time, but also all of the bacteria were eventually killed.

The last graph is shown as no antibiotic given. This is a graph showing the control group. There is always a control group in an experiment where nothing is done to the group. This is used to compare the results in the end of the experiment.

3 0

3 years ago

Blood is forced through the body by the

belka [17]

Blood is pumped throughout the body by your heart. Your heart is the main organ responsible for blood circulation.

8 0

4 years ago

What mass of iron (III) nitrate will be in 129.8ml of a 0.3556 molar aq iron (III) nitrate solution?

Anvisha [2.4K]

Answer: The mass of iron (III) nitrate is 11.16 g/mol

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.3556 M

Molar mass of Iron (III) nitrate = 241.86 g/mol

Volume of solution = 129.8 mL

Putting values in above equation, we get:

$0.3556M=\frac{\text{Mass of iron (III) nitrate}\times 1000}{241.86 g/mol\times 129.8}\\\\\text{Mass of iron (III) nitrate}=11.16g$

Hence, the mass of iron (III) nitrate is 11.16 g/mol

6 0

3 years ago

Argon occupies a volume of 28.3 L at a pressure of 0.36 atm. Find the volume of argon when the pressure is increased to 0.85 atm

FromTheMoon [43]

Answer:

12

Explanation:

Boyles law

P1 (.36)

V1(28.3)

P2(.85)

V2(?)

P1*V1=P2*V2

plug it in, and you get 12

3 0

3 years ago

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:

seropon [69]

Answer: The cell voltage of the given reaction is 1.86 V

Explanation:

The given chemical equation follows:

$3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)$

Oxidation half reaction: $Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V$ ( × 2)

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V$ ( × 3)

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:

$E^o_{cell}=0.16-(-1.66)=1.82V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = $42^oC=[42+273]K=315K$

F = Faraday's constant = 96500

$[Al^{3+}]=1.63M$

$[Cu^{2+}]=3.43M$

Putting values in above equation, we get:

$E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V$

Hence, the cell voltage of the given reaction is 1.86 V

4 0

3 years ago