A substance with a high vapor pressure at normal temperatures is often referred to as volatile. As the temperature of a liquid increases, the kinetic energy of its molecules also increases and as the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.

The balanced redox reaction is

5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)

The unbalanced redox reaction is :

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

The above equation can be balanced  by ensuring the atom of the elements

on the left hand side is equal to those on the right hand side.

We have 3 atoms of Mn on the left side of the equation and 1 atom on the

right hand side.This will be balanced by putting 3 in front of MnO₄− as shown

below:

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 3 atoms of O on the left hand side and 12 atoms of O on the right

hand side. This is  balanced by putting 5 in front of AsO₂− and 2 in front of

H₂O as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 4 atoms of H on the left hand side and 1 atom of H on the right

hand side.We can balance by putting 4 in front of H⁺ as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)

We have 5 atoms of As on the left hand side and 1 atom of As on the right

hand side. We can balance by putting 5 in front of As as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)

The equation is therefore now balanced as the number of atoms of the

element on the left hand side are equal with those on the right hand side.

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Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

Answer:

See explanation below

Explanation:

In both cases the central atoms, C in CHCl₃ and O in H₂O, are sp³ hybridized .

Since they are sp³ hybridized we predict an angle  between the  H-C-Cl and H-O-H of 109.5 º ( tetrahedral ), but two of the sp³ orbitals in water are occupied by lone pairs.

These lone pairs do excercise more repulsion ( need more room ) than the bonds oxygen is making with hydrogen.

As a consequence of this repulsion the angles H-O-H are less than the predicted 109.5º in tetrahedra. ( Actually is 104.5 º)

Answer:

No additional particle was produced during the decay.

Explanation:

The equation of decay is given as;

¹⁰₆C  + ⁰₋₁ e → ¹⁰₅B + x

To identify x, we have to calculate its atomic and mass number.

In the reactants side;

Atomic Number = 6 + (-1) = 5

Mass number = 10 + 0 = 10

In the products side;

Atomic Number = 5 + x

Mass Number = 10 + x

Generally, reactant = product

Atomic Number;

5 = 5 + x

x = 5 - 5 = 0

Mass Number;

10 = 10 + x

x = 10 - 10 = 0

This means no additional particle was produced during the decay.

Explanation:

because T is const so

deltaS=Q/T=nRLn(V2/V1)

=(14/28)x8.314xLn(30/10)=4,567 J/k