Chemical formula for Aluminum Oxide

How much of a 25mg/ml chloramphenicol stock solution do we need to make 400ml of growth media that has final 25ug/ml concentrati

Rama09 [41]

Answer : The volume of chloramphenicol stock solution needed is, 0.4 mL

Explanation :

First we have to calculate the mass of chloramphenicol.

As, the mass of chloramphenicol in 1 mL of solution = $25\mu g=25\times 10^{-6}g$

So, the mass of chloramphenicol in 400 mL of solution = $\frac{400mL}{1mL}\times 25\times 10^{-6}g=0.01g$

Thus, the mass of chloramphenicol is, 0.01 g

Now we have to calculate the volume of chloramphenicol stock solution needed.

Concentration of chloramphenicol = $25mg/mL=25\times 10^{-3}g/mL$

As, $25\times 10^{-3}g$ of chloramphenicol needed 1 mL volume of chloramphenicol stock solution

So, $0.01g$ of chloramphenicol needed $\frac{0.01g}{25\times 10^{-3}g}\times 1mL=0.4mL$ volume of chloramphenicol stock solution

Thus, the volume of chloramphenicol stock solution needed is, 0.4 mL

4 0

3 years ago

Imagine that you helped to discover a new element:

bulgar [2K]

Answer:

Proton = 119

Electron = 119

Neutron = 183

Explanation:

Mass number (A) = 302

Atomic number (Z) = 119

Proton = atomic number = 119

Since there is no charge on the element, it means the element is neutral.

Therefore,

Electron = proton = 119

Mass number = proton + Neutron

Neutron = Mass number — proton

Neutron = 302 — 119

Neutron = 183

7 0

3 years ago

Read 2 more answers

A chlorine Cl and bromine Br atom are adsorbed on a small patch of surface (see sketch at right). This patch is known to contain

MrRissso [65]

Explanation:

It is given that possible number of ways the Cl and Br can be absorbed initially are 100.

S, possible number of ways by which Br can be desorbed is as follows.

$100 \times 99$

Now, we will calculate the change in entropy as follows.

$\Delta S = k_{B} ln (\frac{W}{W_{o}})$

where, $k_{B}$ = Boltzmann constant = $1.38 \times 10^{-23}$

$\Delta S$ = change in entropy

Therefore, we will calculate the change in entropy as follows.

$\Delta S = k_{B} ln (\frac{W}{W_{o}})$

= $1.38 \times 10^{-23} J/K \times ln (\frac{100}{100 \times 99})$

= $1.38 \times 10^{-23} J/K \times -4.595$

= $-6.34 \times 10^{-23} J/K$

Thus, we can conclude that the change in entropy is $-6.34 \times 10^{-23} J/K$ .

5 0

3 years ago

Scientists are studying cold and dry environments on earth that are like the environment on Mars. What kind of prokaryotes do yo

FinnZ [79.3K]

The answer is; extremophiles

These organisms are found in places on earth that would be damaging to most life on earth. There are different categories of extremophiles depending on their environment. For example, those that live in the extreme cold (even below 150C ) are called Psychrophiles while Xerophiles live extreme desiccation.

7 0

3 years ago

Read 2 more answers

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

andre [41]

Complete question:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

Answer:

After 210 s the concentration of SO2Cl2 will be 0.121 M

Explanation:

$ln\frac{[A_t]}{[A_0]} =-kt$

where;

At is the concentration of A at a time t

A₀ is the initial concentration of A

k is rate constant = 1.44×10⁻⁴ s⁻¹

t is time

ln(At/A₀) = -( 1.44×10⁻⁴)t

ln(At/0.125) = -( 1.44×10⁻⁴)210

ln(At/0.125) = -0.03024

$\frac{A_t}{0.125} = e^{-0.03024$

At/0.125 = 0.9702

At = 0.125*0.9702

At = 0.121 M

Therefore, after 210 s the concentration of SO2Cl2 will be 0.121 M

6 0

3 years ago