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Greeley [361]
4 years ago
9

What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?

Chemistry
1 answer:
frutty [35]4 years ago
5 0

Answer:

6.93

Explanation:

Step 1: Given data

  • Standard Gibbs free energy (∆G°): -5.20 kJ
  • Temperature (T): 50°C
  • Equilibrium constant (K): ?

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

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ser-zykov [4K]

Answer:: Mendel studied how traits are been passed from parents to offspring using seven features in peas, including height, flower color, seed color, and seed shape. To do this he divided the pea plant into short height and tall height. From this experiment he proposed a principle called independent assortment, which describes how different genes independently separate from one another when reproductive cells develop. Though this experiment was studied using gene formation in prokaryotic cell.

This principle of independent assortment is also seen in eukaryotic cells during meiosis.

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4 0
4 years ago
When 2.50 g of an unknown weak acid (ha) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of
baherus [9]
When dT = Kf * molality * i
                = Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
                               = 2.5g /250g x 1 mol /85 g x1000g/kg
                               =0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution 
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and  i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
     = (0.1176 * 0.359)^2 / (1-0.359)
     = 2.8x10^-3



5 0
3 years ago
Consider the following reversible reaction.
andriy [413]

Answer:

No one is correct. The correct expression is:

Keq = [H₂]²  . [O₂]² / [H₂O]²

Explanation:

To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.

The balance reaction is:

<u>2</u> H₂O (g)  ⇄  <u>2</u> H₂ (g)  +  O₂ (g)

Keq = [H₂]²  . [O₂]  / [H₂O]²

In opposite side: <u>2</u> H₂ (g)  +  O₂ (g)   ⇄  <u>2</u> H₂O (g)

Keq =  [H₂O]² / [H₂]²  . [O₂]  

6 0
4 years ago
Calculate the mass of water produced when 2.06 g of butane reacts with excess oxygen.
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1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10

2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O

3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O
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7 0
3 years ago
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