That statement may be considered true. 'Specific gravity' is the ratio of
the density of a substance to the density of a reference substance. The
reference substance is nearly always water, and the ratio of the density
of water to the density of water is obviously ' 1'. So there you are.
Answer:the correct answer is 1.0* 10^-7 mol/L
Explanation:
Answer:
T_2= 234.37 K
Explanation:
According to Claperyon, we know that

P_1= Atmospheric pressure 760 mm Hg
P_2 = pressure at the bottom of the column
= 10×10^3 mm of Hg+ 760 mm of Hg
= 10760 mm of Hg
now,
P_2-P_1= 10760-760= 10^4 mm
P_2-P_1 ( in pascals) = 10^4× 133.322= 1333220 mm
the enthalpy of fusion (ΔH-fus) of mercury is 2.292 KJ/mol
use the above equation to calculate ΔT as follows

therefore, T_2= 234.37 K
Answer:
True, applied force is the force of support exerted by an object that holds up another object.
Answer:
i cant solve this!
Mybe i can solve another question!