4. 200 ml of a 5.6m solution to bacl_2 have 750 ml of water added to it. what is the new concentration?

14. the gas laws can also be expressed graphically. sketch the following plots for an ideal gas. (see picture)

4. 1.18 mol·L⁻¹

14. See below.

Explanation:

4. Dilution calculation

V₁c₁ = V₂c₂

Data:

V₁ = 200 mL; c₁ = 5.6 mol·L⁻¹

V₂ = 950 mL; c₂ = ?

Calculation:

c₂ = c₁ × V₁/V₂

c₂ = 5.6 mol·L⁻¹ × (200/950) = 1.18 mol·L⁻¹

The new concentration is 1.18 mol·L⁻¹ .

14. Boyle's Law graphs

We can write Boyle's Law as

pV = k or p = k/V or V= k/p

p and V are inversely related.

(a) As pressure increases, volume decreases. Thus, a graph of V vs p is a hyperbola.

(b) p = k/V =k(1/V)

 1/V = (1/k)p

   y  =   m x  + 0

A graph of 1/V vs p is a straight line.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

it is a solid

explanation: