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Bas_tet [7]
2 years ago
14

Plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me

Chemistry
1 answer:
larisa [96]2 years ago
6 0

Answer:

Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2  x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A  and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

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3 0
3 years ago
Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.
SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

                HClO   ⇄   ClO-  +  H+

start:          0.05            0         0

change       -x               +x       +x

balance     0.05-x         x         x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

3x10^{-8}=\frac{x^{2} }{0.05}

clearing the x and calculating its value we have:

x=3.87x10^{-5}=[H+]=[ClO-]

the pH can be calculated by:

pH=-log[H+]=-log[3.87x10^{-5}]=4.41

7 0
3 years ago
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7 0
3 years ago
Read 2 more answers
During a lab, you heat 1.62 g of a CoCl2 hydrate over a Bunsen burner. After heating, the final mass of the dehydrated compound
sashaice [31]
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
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n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
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m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O) 
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
8 0
3 years ago
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