For the Haber process,N2+3H2 yields 2NH3, what volume of Nitrogen is consumed at STP if you collect 44.8 L of ammonia in excess hydrogen (N=14 AMU, H=1 AMU)a. 14.9 L
b. 29.9 L
c. 44.8L
d. 22.4 L

Answer:

The gas above it or boiling

Explanation:

Answer:

70g

Explanation:

Rate = 35g of NaCl can dissolve in 100g of H2O

Use proportions for 200g of H2O:

Since 200 is double of 100, x needs to be double of 35.

Therefore the maximum amount of NaCl that can dissolve in 200g of water is 70g.

Answer: Option (b) is the correct answer.

Explanation:

In liquids, the molecules are held by strong intermolecular forces of attraction as compared to gases. Due to which they are able to collide and slide past each other. Hence, they have a medium kinetic energy.

And, in liquid molecules there is moderate intermolecular distance between the particles.

In gases, the molecules are held by weak Vander waal forces. Hence, they have high kinetic energy due to which they move very rapidly from one place to another leading to more number of collisions.  

Hence, gases are able to expand more rapidly as compared to liquids. Therefore, between the atoms of a  gas there are large intermolecular distances.

Therefore, we can conclude that the statement particles collide in gas so Substance X is a gas, best describes the two substances.

your answer is Particles collide in gas so Substance X is a gas.

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

Final answer:

The lattice energy for rubidium chloride (RbCl) is calculated by substituting the given values into the equation derived from Hess's Law. The calculated lattice energy is found to be -695 kJ/mol.

Explanation:

In this question, you are asked to select the lattice energy for rubidium chloride (RbCl). The lattice energy can be calculated using various given energies including enthalpy of formation (ΔHf), electron affinity (Cl), enthalpy of sublimation, ionization energy, and bond dissociation energy. Using Hess's Law, this can be summed up as:

ΔHf(RbCl) = [Sublimation Energy (Rb) + Ionization Energy (Rb) + 0.5 × Bond Energy (Cl₂) + Electron Affinity (Cl)] - Lattice Energy (RbCl)

By rearranging this formula, we find that the Lattice Energy (RbCl) = [Sublimation Energy (Rb) + Ionization Energy (Rb) + 0.5 × Bond Energy (Cl₂) + Electron Affinity (Cl)] - ΔHf(RbCl). Substituting in the given values, we find the lattice energy to be -695 kJ/mol. Therefore, the correct option is C. -695.

Learn more about Lattice Energy here:

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