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nasty-shy [4]
2 years ago
10

In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] = 1.3

M. What is the equilibrium constant Kc for this reaction?
Chemistry
1 answer:
xenn [34]2 years ago
5 0

Answer: The equilibrium constant K_c for this reaction is 0.088

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For the given chemical reaction:

NO+NO_2(g)\rightarrow N_2O_3

The expression for K_c is written as:

K_c=\frac{[N_2O_3]}{[NO_2][NO]}

K_c=\frac{(1.3)}{(3.9)\times (3.8)}

K_c=0.088

The equilibrium constant K_c for this reaction is 0.088

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Carbon dating of small bits of charcoal used in cave paintings has determined that some of the paintings are from 10000 to 30000
Andru [333]

The age of painting was determined from the decay kinetics of the radioactive Carbon -14 present in the painting sample.

Given that the half life of Carbon-14 is 5730 years.

Radioactive decay reactions follow first order rate kinetics.

Calculating the decay constant from half life:

λ= \frac{0.693}{t_{1/2} }

        = \frac{0.693}{5730 yr} = 1.21*10^{-4}yr^{-1}

Setting up the radioactive rate equation:

ln\frac{A_{t} }{A_{0} } =-kt

Where A_{t} = Activity after time t = 0.80microCi

A_{t} = initial activity = 6.4microCi

k = decay constant = 1.21*10^{-4}yr^{-1}

ln\frac{0.80uCi}{6.4uCi} =-(1.21*10^{-4}yr^{-1})t

ln 0.125 = -(1.21*10^{-4}yr^{-1})t

-2.079=-(1.21*10^{-4}yr^{-1})t

t=\frac{2.07944}{1.21*10^{-4} } yr

 = 17185 years

t = 17185 years

Therefore age of the painting based in the radiocarbon -14 dating studies is 17185 years



6 0
2 years ago
The double bond between carbon and oxygen is similar to an alkene C-C, except that C o is: a) shorter and weaker. b) shorter and
mina [271]

Answer:

the double bond between c and o is shorter and weaker

Explanation:

this is because the bond between c and o involves unequal sharing of electrons whole c and c involves hybridization sp2 of orbitals and also catenation phenomenon in which carbon could form long chain with it's other carbon

5 0
2 years ago
A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?
Xelga [282]

let the actual height of car be x

now, according to question,

  • \dfrac{1}{15}  =  \dfrac{8}{x}

  • x = 15 \times 8

  • x = 120 \:  \: cm

  • height \:  \: of \:  \: car   =   120 \:   cm \:  \: or \:  \: 1.2 \:  \: m
5 0
1 year ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
Compound w , c6h13cl, undergoes base-promoted e2 elimination to give a single c6h12 alkene, y. compound x, c6h13br, undergoes a
CaHeK987 [17]
<span>Answer: W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation: X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>
5 0
2 years ago
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