__________ force is the force of attraction and repulsion between subatomic particles.

Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.

Hatshy [7]

Following reaction occurs in the given electrochemical system:
$Fe^{2+} + Zn$ → Fe + $Zn^{2+}$
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where, $E^{0}Fe2+/Fe$ = standard reduction potential of Fe2+/Fe = -0.44 v
$E^{0}Zn2+/Zn$ = standard reduction potential of Zn2+/Zn = -0.763 v

E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, Δ $G^{0} = -RTln(K) ............ (2)$

On equating and rearranging equation 1 and 2, we get
K = exp( $\frac{nFE(0)}{RT}$ )= exp ( $\frac{2X96500X0.323}{8.314X298}$ ) = 8.46 x $10^{10}$

7 0

2 years ago

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

strojnjashka [21]

Answer:

Force of attraction = 35.96 $\times 10^{27}$ N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = $10^{-9}$ m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$ N

6 0

2 years ago

If 0.55 g of a gas dissolved in 1.0 l of water at 20.0 kpa of pressure, how much will dissolve at 110.0 kPa of pressure?

GaryK [48]

Answer:sup

Explanation:

4 0

3 years ago

Is sulfur a metal, nonmetal, metalloid, or noble gas?

Blababa [14]

Sulfer is a nonmetal

6 0

2 years ago

Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how

Aleonysh [2.5K]

Given equation : $5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O$

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

$Moles = \frac{grams}{molar mass}$

Molar mass of KNO2 = 85.10 g/mol

$Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}$

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

$0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}$

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

$Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}$

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

$2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}$

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0

3 years ago