Answer:
Drug calculation
If we have 45g of clobetasol = 0.05%w/w
Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g
It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w
Answer:
c = 4016.64 j/g.°C
Explanation:
Given data:
Mass of substance = 2.50 g
Calories release = 12 cal (12 ×4184 = 50208 j)
Initial temperature = 25°C
Final temperature = 20°C
Specific heat of substance = ?
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Solution:
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 20°C - 25°C
ΔT = -5°C
50208 j = 2.50 g . c. -5°C
50208 j = -12.5 g.°C .c
50208 j /-12.5 g.°C = c
c = 4016.64 j/g.°C
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m
where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m
</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC
Now, for water freezing point = 0 oC
∴Freezing point of solution = -2.604 oC
Answer:
100 cg/1g
Step-by-step explanation:
1 cg = 0.01 g Multiply by 100
100 cg = 1 g
(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.
(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.
(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.
