Which statements are wrong about atoms

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

Carbon dioxide and helium which has a higher van der waals force of attraction between molecules

raketka [301]

Answer:

Carbon dioxide

Explanation:

Neither helium nor carbon dioxide has a molecular dipole, so their strongest van der Waals attractive forces are London forces.

Helium is a small spherical atom with only a two electrons, so its atoms have quite weak attractions to each other.

CO₂ is a large linear molecule. It has more electrons than helium, so the attractive forces are greater. Furthermore, the molecules can align themselves compactly side-by-side and maximize the attractions (see below).

For example. CO₂ becomes a solid at -78 °C, but helium must be cooled to -272 °C to make it freeze (that's just 1 °C above absolute zero).

7 0

3 years ago

The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago

Chalcopyrite is an ore with the composition cufes2. what is the percentage of iron in a 39.6 g sample of this ore? answer in uni

iren2701 [21]

<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>

5 0

3 years ago

If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is

Cloud [144]

Answer:

Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.

Explanation:

The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:

Time_________________ Amount

t=0_____________________x

t=10,000 years____________x/2

t=20,000 years___________x/4

During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

6 0

3 years ago