Answer:
The main advantage would be that with the pouring temperature being much higher, there is very little chance that the metal will solidify in the mould while busy pouring. This will allow for moulds that are quite intricate to still be fully filled. The drawbacks, though, include an increased chance defects forming which relates to shrinkage (cold shots, shrinkage pores, etc). Another drawback includes entrained air being present, due to the viscosity of the metal being low because of the high pouring temperature.
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer:
The sun'll likely absorb the radiation if it is close enough (Which it will never be)
Explanation:
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:
P1V1 = P2V2,
Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:
(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr
The final pressure exerted by the gas would be 0.593 torr.
Hope this helps!
Answer:
The minimum concentration of acetaminophen that can be detected by new= 10μg/mL
Volume of blood sample=2ml
Minimum mass of acetaminophen that can be detected by automated system= 10×2= 20μg= 0.020mg
