Which pair could be genetic clones? look at the photo

CHEMISTRY HELP! Lewis structure rules Please help, is this correct?

fiasKO [112]

Answer:

it is the one below that. NO, because it debt net the octet rule

4 0

3 years ago

If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of

valkas [14]

Answer: Osmolarity of the sodium chloride solution is 18 Osmol/L.

Explanation:

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

Osmole is defined as the particles that contribute to the osmotic pressure of a solution.

The solute given here is sodium chloride (NaCl). Number of osmoles can be determined by the dissociation of this solvent into ions.

The equation given by the dissociation of NaCl:

$NaCl\rightarrow Na^++Cl^-$

1 mole of sodium chloride produces 2 moles of ions.

So, 3 moles of sodium chloride will produce = (3 × 2) = 6 moles of ions.

Hence, osmolarity of the sodium chloride solution will be $6\times 3mol/L=18Osmol/L$

5 0

3 years ago

Which of the following best compares the structure of carbohydrates and nucleic acids?

Serga [27]

I believe the best description comparing the structure of carbohydrates and nucleic acids would be A.

5 0

3 years ago

Read 2 more answers

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0

3 years ago

Which of the following is an example of a distance force?

elixir [45]

C. A magnet causing a metal object to move

4 0

3 years ago