How many molecules are in 2.64 moles of Na2O?

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Below are the choices that can be found elsewhere:

a. [CH3NH-] * [OH-] / [CH3NH3+]
b. [CH3NH3+] / ([CH3NH2] * [OH-]) 
c. [CH3NH3+] * [OH-] / [CH3NH2] 
d. [CH3NH3+] * [OH-] / [CH3NH2] * [H2O]

The answer is C.

4 0

3 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

Shtirlitz [24]

Answer:

(a) $pH = -Log (0.1M) = 1$

(b) $pH = -Log (10^{-13}M) = 13$

(c) $pH = -Log (3x10^{-3}M) = 2.5$

(d) $pH = -Log (4.93x10^{-10}M) = 9.3$

(e) $pH = -Log (5^{-7}M) = 6.3$

Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$

clearing the $[H^{+} ]$

$[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}$

(d) is a weak base so it is necessary to solve the equilibrium first, knowing $Ka=3.24x10^{-8}$

The reaction is $HClO$ → $H^{+} + CO^{-}$ so the equilibrium is

$Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}$

clearing the x

${x^{2}={1.62x10^{-17}-3.24x10^{-8}x}$

$x=[H^{+}]=4.93x10^{-10}$

6 0

3 years ago

Suppose you decide to define your own temperature scale using the freezing point (13 degrees C) and boiling point (360 degrees C

podryga [215]

M.P of oleic acid = 13 deg C
B.Pof oleic acid = 360 deg C 

(360 - 13) / 100 = 3.47 
So 1 deg O = 3.47 deg C 

The scale do not start at 0 deg C, it starts at 13 deg C. So to convert deg C to deg O,
subtract 13 then divide by 3.47 

deg O = (deg C - 13) / 3.47 

convert O to C multiply by 3.47 then add 13 

deg C = (deg O x 3.47) + 13 

convert 0 deg C to deg O 
deg O = (0 deg C - 13) / 3.47 
= - 3.75 deg O

8 0

3 years ago