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Ainat [17]
3 years ago
13

What is the first way in which biology researchers present the results of their latest research?

Chemistry
1 answer:
antoniya [11.8K]3 years ago
5 0
Almost always, the first way in which biology researchers present the results of their latest research is to discuss the results of previous research that they want to build off of. 
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I just need help on c-g. The whole question is there in case anyone needs it.
Elza [17]

Answer:

(c) 0.11; (d) -24.5 kJ·mol⁻¹; (e) See below; (f) Kc increases;

(g) No effect on ΔH

Step-by-step explanation:

(c) Kc at 125 °C  

In Part (a) one molecule of XY dissociated into one X and one Y.

       XY ⇌ X + Y

I:      10       0    0

C:     -1       +1   +1

E:      9        1     1

Kc = {[X][Y]}/[XY] = (1 × 1)/9 = ⅑ = 0.11

(d) ΔH

ΔH is a constant that is characteristic of the reaction.

ΔH = -24.5 kJ·mol⁻¹

(e) Effect of temperature on concentrations

If ΔH is negative, the reaction is exothermic.

Heat is a product of the reaction, so we can write the equation as

XY ⇌ X + Y + heat

If we lower the temperature, we are removing heat from the system.

<em>Le Châtelier's Principle</em> states that if you apply a stress to a system at equilibrium, it will respond by trying to relieve the stress.

We applied a stress by removing heat, so the system responds by producing more heat. The position of equilibrium moves to the right, and <em>more products will form</em>.

The diagram might look like the one below.

(f) Effect of temperature on Kc

Kc = [Products]/[Reactants]

We are increasing [Products] and decreasing [Reactants].

If you increase the numerator and decrease the denominator, you i<em>ncrease</em> the value of the quotient.

The value of the equilibrium constant increases when the temperature decreases.

(g) Effect of temperature on ΔH

Decreasing the temperature has no effect on ΔH, because the enthalpies of the reactants and products are properties of the substances themselves. They do not depend on the temperature.

7 0
2 years ago
Consider the reaction FeO (S) + CO(g) &lt;-----&gt; Fe(s) + CO2(g) for which KP is found to have the following values:
Svetach [21]

Explanation:

\Delta G^o=-RT\ln K_1

where,

R = Gas constant = 8.314J/K mol

T = temperature = 600^oC=[273.15+600]K=873.15 K

K_p = equilibrium constant at 600°C =  0.900

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )

\Delta G^o=764.85 J/mol

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = K_1=0.900

Equilibrium constant at 1000°C = K_2=0.396

T_1=[273.15+600]K=873.15 K

T_2=[273.15+1000]K=1273.15 K

\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]

\Delta H^o=-18,969.30 J/mol

The ΔH° of the reaction at 600 C is -18,969.30 J/mol.

ΔG° = ΔH° - TΔS°

764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°

ΔS° = -22.60 J/K mol

The ΔS° of the reaction at 600 C is -22.60 J/K mol.

FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)

Partial pressure of carbon dioxide = p_1=P\times \chi_1

Partial pressure of carbon monoxide = p_2=P\times \chi_2

Where \chi_1\& \chi_2 mole fraction of carbon dioxide and carbon monoxide gas.

The expression of K_p is given by:

K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}

0.900=\frac{\chi_1}{\chi_2}

\chi_1=0.900\times \chi_2

\chi_1+\chi_2=1

0.9\chi_2+\chi_2=1

1.9\chi_2=1

\chi_2=\frac{1}{1.9}=0.526

\chi_1=1-\chi_2=1-0.526=0.474

Mole fraction of carbon dioxide at 600°C is 0.474.

6 0
3 years ago
Why does sodium carbonate affect pH?
Varvara68 [4.7K]
Carbonate compounds have a negative valency (because they contain oxygen), so NaCO3 has a negative valency (it’s a negatively-charged ion) which makes it basic or alkaline.
8 0
3 years ago
Why does his teacher ask him to balance the equation by including the correct coefficients?
goldenfox [79]
Because correct coefficients are necessary to successfully balance equation.
8 0
3 years ago
Definition: this is the amount of disorder in a system
Sholpan [36]

Answer:

Entropy

Explanation:

4 0
2 years ago
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