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kramer
3 years ago
14

g The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures in excess of 1100 K. We can conclude that a. ΔG° is ne

gative for all temperatures. b. ΔH° is negative and ΔS° is negative. c. ΔH° is positive and ΔS° is positive. d. ΔH° is negative and ΔS° is positive. e. ΔH° is positive and ΔS° is negative.
Chemistry
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

c. ΔH° is positive and ΔS° is positive.

Explanation:

Hello,

In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

\Delta G=\Delta H-T\Delta S

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.

Best regards.

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A box contains identical balls of which 12 are red, 18 white and 8 blue. Three balls are drawn from the box one after the other
svet-max [94.6K]

Answer:

a) 12/323

b) 8/233

Explanation:

a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323

b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233

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3 years ago
Which of these groups includes organisms that are MOST closely related?
gregori [183]
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If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

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6 0
3 years ago
True or false?
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3 0
3 years ago
determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)
sasho [114]

The  mass  of water formed  is


<u><em>calculation</em></u>

Use  the  ideal   gas  equation   to  calculate the  moles of  NH3  and O2

that  is  Pv= n RT

where;  P= pressure,  

V=  volume,

n = number  of  moles,

R=gas   constant  = 0.0821  l .atm/ mol.K

make n the formula of  the subject  by diving   both side  by  RT

n =  PV /RT

The   moles of NH3

n= (1.50 atm  x 12.5 L) /(  0.0821 L. atm /mol.k   x 298 K)  =0.766  moles

The  moles  of  O2

=(1.1 atm  x 18.9  L) /  (  0.0821 L. atm/ mol.k   x 323 K) = 0.784  moles


write the reaction  between  NH3  and  O2

4 NH3  + 5 O2  →4 No  +6H2O


from  equation above  0.766  moles of NH3  reacted to produce  

0.766 x 6/4 =1.149 moles of H2O


0.784  moles of O2   reacted to  produce  0.784  x 6/5=0.9408  moles  of H20


since  O2  is totally  consumed, O2  is the limiting  reagent  and therefore  the  moles of H2O  produced=  0.9408  moles


mass  of  H2O  = moles x molar mass

 from  periodic table the  molar mass  of H2O  =  (1 x2)+16= 18  g/mol

mass = 18 g/mol  x 0.9408  moles= 16.93  grams


3 0
3 years ago
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