Molality is a way to express the concentration of a solution that represents the number of moles of

What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?

AlekseyPX

I think c ... C this is the correct

8 0

3 years ago

Valence electrons<br> Fill in the blanks

IgorC [24]

Answer:

Valence electrons or outer electrons are most important as they participate in bonding. The octet rule states that atoms gain, lose, or share valence electrons to have filled energy levels.. this gives atoms a stable configuration like that of the nearest noble gas.

6 0

3 years ago

A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin

Illusion [34]

Answer:

6He = 4.90 MeV/Nucleon

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its protons and neutrons, this mass is greater than the actual value . This is the mass defect.

Now this mass defect we can convert to energy by utilizing Einstein´s equation, E = mc².This is the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons = 2 x 1.0078 u = 2.0516 u

mass neutrons = 4 x 1.0087 u = 4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹² J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

8 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago

Define significant figure. Discuss the importance of using the proper number of significant figures in measurements and calculat

anastassius [24]

Explanation:

Significant figure is the measure of how accurately something can be measured. It carries meaning contributing to its measurement resolution. It is important to use proper number of significant figures to get a precise measurement. For example, if we use a meter stick then measurements like 0.874 meters, or 0.900 meters, are good because they indicate that we can measure to the nearest millimeter. Whereas a measurement like 0.8 does not tell that a meter stick can measure to the nearest millimeter.

7 0

3 years ago