Bacterial cells can pick up the DNA through the process of transformation. Lambda phages are used by eliminating the middle of its liner genome and adding the foreign DNA in the created space. The phage is then introduced into the bacterial cell where it replicated itself via the lysogenic cycle.

8 0

3 years ago

If you stay at 90 degrees Celsius and a gas in a liquid solution has a pressure of 2 atm, what solubility should it have if it h

grigory [225]

Answer:

im not sure

Explanation:

5 0

3 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago

Determine the oxidation number of sodium in Na202

Olegator [25]

Answer:

Explanation:

Na₂O₂

NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.

Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

2Na + 2O = 0

O = –1

2Na + 2(–1) = 0

2Na – 2 = 0

Collect like terms

2Na = 0 + 2

2Na = 2

Divide both side by 2

Na = 2/2

Na = +1

Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1

5 0

3 years ago