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3 years ago

How does molecular attraction affect the amount of energy needed to create a phase change. Think about low, medium, and high

Chemistry

1 answer:

Art [367]3 years ago

5 0

Answer:

When considering phase changes, the closer molecules are to one another, the stronger the intermolecular forces. Good! For any given substance, intermolecular forces will be greatest in the solid state and weakest in the gas state.

In the case of melting, added energy is used to break the bonds between the molecules. ... If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

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What will most likely happen when stress is applied to an equilibrium reaction? The system will not respond to the stress. The s

steposvetlana [31]

Answer:

The system will change its concentration to shift to a new equilibrium position.

Explanation:

For example in the Haber Process

N2 + 3H2 ⇄ 2NH3

If the pressure is increased the process will move to the right - to have more NH3 and less of the nitrogen and hydrogen.

8 0

3 years ago

Read 2 more answers

Use the definition of molarity to calculate the concentration of 12.34 g of CaSO4 completely dissolved in water, with a solution

Ede4ka [16]

Answer:

[CaSO₄] = 36.26×10⁻² mol/L

Explanation:

Molarity (M) → mol/L → moles of solute in 1L of solution

Let's convert the volume from mL to L

250 mL . 1L/1000 mL = 0.250L

We need to determine the moles of solute. (mass / molar mass)

12.34 g / 136.13 g/mol = 0.0906 mol

M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L

8 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$