<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:

Or,

where,
= 
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:

Hence, the freezing point of solution is 2.6°C
Answer:
Explanation:
The wavelength is the distance between two adjacent wavefronts. ... If the wave crosses to the new medium at an angle (not 90 degrees), the change ... When light enters a more optically dense medium, it is refracted closer to the normal. the same as the critical angle, light will travel along the boundary of the 2 mediums.
Explanation:
Haemoglobin consists of heme unit which is comprised of an <u>
</u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.
The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>
The organic ring component of heme is - <u>Porphyrin</u>
Under normal conditions, the central atom of heme is - <u>
</u>
In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.
C. Rubbing the balloon against your hair
Answer: 72L of 30% and 128L of 80%
You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1
The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L
v1+v2=200L
v2= 200L- 72L= 128L