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3 years ago

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What may be expected when k > 1.0?

1 answer:

inessss [21]3 years ago

8 0

When k is greater than 1, this means the equilibrium concentration of the products is greater than the equilibrium concentration of the reactants. K is expressed as the equilibrium concentration of the products divided by the <span>equilibrium concentration of the reactants. This also means the forward reaction is favored. </span>

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According to vsepr theory, a molecule with the general formula ax4e2 will have a _____ molecular shape. easynotes

Lena [83]

According to Vsepr theory, a molecule with the general formula ax4e2 will have a square planar molecular shape.The molecule of this geometry has their atoms which are positioned at the corners of the square on the same plan about a central atom.

7 0

3 years ago

BartSMP [9]

Answer:

Odds to be given for an event that either Romance or Downhill wins is 11:4

Explanation:

Given an odd, r = a : b. The probability of the odd, r can be determined by;

Pr(r) = $\frac{a}{b}$ ÷ (

So that;

Odd that Romance will win = 2:3

Pr(R) = $\frac{2}{3}$ ÷ (

= $\frac{2}{3}$ ÷ $\frac{5}{3}$

= $\frac{2}{5}$

Odd that Downhill will win = 1:2

Pr(D) = $\frac{1}{2}$ ÷ (

= $\frac{1}{2}$ ÷ $\frac{3}{2}$

= $\frac{1}{3}$

The probability that either Romance or Downhill will win is;

Pr(R) + Pr(D) = $\frac{2}{5}$ + $\frac{1}{3}$

= $\frac{11}{15}$

The probability that neither Romance nor Downhill will win is;

Pr(neither R nor D) = (1 - $\frac{11}{15}$ )

= $\frac{4}{15}$

The odds to be given for an event that either Romance or Downhill wins can be determined by;

= Pr(Pr(R) + Pr(D)) ÷ Pr(neither R nor D)

= $\frac{11}{15}$ ÷ $\frac{4}{15}$

= $\frac{11}{4}$

Therefore, odds to be given for an event that either Romance or Downhill wins is 11:4

8 0

2 years ago

Easy Chemistry/ Help

Sauron [17]

I’m not really sure what the qn is asking for but if i’m not wrong it should be ‘1’ for each blank!

4 0

3 years ago

The blackbody curve for a star named beta is shown below. The most intense radiation for this star occurs in what spectral band?

Viefleur [7K]

The answer is Infrared.

5 0

2 years ago

Read 2 more answers

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

taurus [48]

Answer:

<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

Explanation:

Adding solute to water causes the depression of the freezing point.

We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

<em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

6 0

3 years ago