The compound methane can be broken down into carbon and<br> hydrogen<br> TRUE<br> FALSE

Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will

Alla [95]

Answer:

Only $Fe^{3+}+Mg$ gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential ( $E^{0}$ ) of the reaction is positive. Because it leads to negative standard gibbs free energy change ( $\Delta G^{0}$ ), which is a thermodynamic condition for spontaneity of a reaction.

$E^{0}=E^{0}(reduction)-E^{0}(oxidation)$

Where $E^{0}(reduction)$ and $E^{0}(oxidation)$ represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation: $Mg-2e^{-}\rightarrow Mg^{2+}$ ; $E_{Mg^{2+}\mid Mg}^{0}=-2.38V$

Reduction: $Fe^{3+}+3e^{-}\rightarrow Fe$ ; $E_{Fe^{3+}\mid Fe}^{0}=-0.04V$

So, $E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V$

Hence this pair will give spontaneous reaction.

(2) Similarly as above, $E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V$

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, $E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V$

Hence this pair will give non-spontaneous reaction.

(4) Similarly as above, $E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V$

Hence this pair will give non-spontaneous reaction.

6 0

3 years ago

Which of the following observations about pH is consistent for an acidic solution?

kow [346]

Answer:

A pH value of 3

Explanation:

The pH scale has 14 numbers:

- 7 is a true natural

-anything above 7 is a base

-and anything under 7 is an acid

pH of three is a strong acid

5 0

1 year ago

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

3 years ago

To name the compound written as CuCl2, you would write:

vovangra [49]

Answer: option C. Copper (II) chloride

Explanation:

To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:

Cu + 2Cl = 0 (since the compound has no charge)

Cl = —1

Cu + 2(—1) = 0

Cu —2 = 0

Collect like terms

Cu = 0 +2

Cu = +2

Therefore, the oxidation state of Cu in CuCl2 is +2.

The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.

4 0

3 years ago

PLEASE HELP QUICK AS POSSIBLE!!!!

GalinKa [24]

Answer:

4180J

Explanation:

(25.0g)(4.184J/g°C)(75°C-35.0°C)

(25.0g)(40.0°C)(4.184J/g°C)

(1.00*10³g°C)(4.184J/g°C) = 4184J

use sig figs:

4180J

8 0

3 years ago

The compound methane can be broken down into carbon and hydrogen TRUE FALSE