A chemical reaction that ______ energy can be used to heat up other substances.

What attracts heat more? black or aluminum foil?

Inessa05 [86]

Answer:

"ALUMINUM FOIL" will conduct heat much more effectively than black construction paper.

8 0

3 years ago

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If a solution containing 0.10 M Cl-, Br-, I-, and LaTeX: CrO_4^2-C r O 4 2 − is treated with Ag+, in what order will the anions

Musya8 [376]

Answer:

$AgI \longrightarrow AgBr \longrightarrow Ag_2CrO_4 \longrightarrow AgCl$

Explanation:

The first thing to have in mind is that the Ksp (constant of solubility) gives you an idea of how much soluble a salt is. If the concentration of ions of that salt exceeds the kps, the salt will precipitate.

The Kps of all these salts are very small compared to the 0.1 M concentration of the anions so all of them will precipitate.

The smaller the constant, the faster the salt precipitates

Having that concept in mind, the order of precipitation from first to last would be:

$AgI \longrightarrow AgBr \longrightarrow Ag_2CrO_4 \longrightarrow AgCl$

6 0

3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M