A) Ca(OH)2 + CO2 —> CaCO3 + H2O
B) when Ca(OH)2 is reacted with CO2, the CaCO3 produced is a precipitate which turns the solution milky
Answer:
The molar mass of the unknown gas is 
Explanation:
Let assume that the gas is O2 gas
O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.
Under the same conditions;
the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.
From Graham's Law of Diffusion;
Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.
i.e
where K = constant
If we compare the rate o diffusion of two gases;
Since the density of a gas d is proportional to its relative molecular mass M. Then;
Rate is the reciprocal of time ; i.e
Thus; replacing the value of R into the above previous equation;we have:
We can equally say:
Answer: 362,07 cm3
To answer this question you need to convert the lb into gram first. One lb equal to 453.592g, so: 3.6lb x 453.592gram/lb= 1632.9312gram.
Now we have mass(1632.9312g), density (4.51g/cm3). Volume is mass divided by density. The equation would be:
Volume= mass/density
Volume = 1632.9312gram / (4.51g/cm3)= 362,07 cm3
Answer:
1.772 gram is the approximate answer
Explanation:
molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole
the reaction is
AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl
from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3
implies 132g AlCl3 gives 78g Al(OH)3
Implies 3g AlCl3 gives
3*122/78 = 1.772 grams
Answer:
92gm
Explanation:
Atomic mass of Mg=24g=1 mole of Mg
∴ 24g =1 mole of Mg contain 6.022×10^23 atom
∴ 6gm contains 246.022×1023×6
=4×6.022×10^23 atoms
Now according to question, there are 6.022×1023 atoms of Na
23gm of Na contains 6.022×10^23 atoms
∴6.022×4×10^23 atoms of Na weighs 23×6.022×10^23×4/6.022×10^23⇒92gm
