What term is best described as the rate of a chemical reaction at any instant in time?

How many moles of water are produced when 3.45 moles of KMnO4 react? Give your answer to the nearest 0.1 moles. moles H2O

andrew11 [14]

Answer:

13.8 moles of water produced.

Explanation:

Given data:

Moles of KMnO₄ = 3.45 mol

Moles of water = ?

Solution:

Chemical equation:

16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Moler ratio of water and KMnO₄:

KMnO₄ : H₂O

2 : 8

3.45 : 8/2×3.45 = 13.8 mol

Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.

7 0

3 years ago

At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rat

Svetllana [295]

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

5 0

3 years ago

Jamie rolls a 6-sided die 30 times and determines that the experimental probability of rolling a 2 is Start Fraction 1 over 15 E

anyanavicka [17]

Answer:

Conduct more trials

Explanation:

Theoretical Probability can be defined as what someone is expecting to happen

Experimental Probability on the other hand, is defined as what actually happens.

Probability is usually calculated in the same way for experimental probability and that of theoretical probability. You divide the total number of possible ways in which a particular outcome can happen, by the total number of outcomes itself.

In Experimental probability, the more times a probability is tried, it gets closer and even more closer to theoretical probability.

So, for the question, Jamie should improve the number of tries more, so as to get his experimental probability results to be closer to the theoretical probability result.

8 0

3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

4 0

3 years ago

Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete

neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = $-RTInK_(eq)$

where:

R= universal gas constant

T= temperature

$K_eq$ = equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = $-RTInK_(eq)$

where;

[texK_eq[/tex]= $\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}$

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

$H^+_{inside}$ ⇒ $H^+_{outside}$

Equilibrum constant for the transport is given as:

$K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}$

$=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}$

$[H^+]_{cell}$ = 10⁻⁷⁴

=3.98 * 10⁻⁸M

$[H^+]_{stomach lumen}$ = 10⁻²¹

=7.94 * 10⁻³M

Hence;

$K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}$

= $\frac{3.98*10^{-8}}{7.94*10{-3}}$

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = $-RTInK_(eq)$

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = $-(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})$

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = $-nFE°_{membrane}$

ΔG₂ = $-(1 mol)(96.5KJ/mol/V)(60*10^{-3})$

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ $G_{total} = G_{1}+G_{2}$

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0

3 years ago