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rewona [7]
3 years ago
8

Select the compound with the lowest lattice energy.

Chemistry
1 answer:
tatyana61 [14]3 years ago
6 0

Answer:

A. CsBr(s)

Explanation:

we will get here compound with the lowest lattice energy

solution

As we know that Lattice energy is always proportional to the charge of ions and it is inversely proportional to the size of ions.

so that by the smallest charge and the largest size give us the lowest lattice energy and that charge and size is express as here as

Charge

Cs (+1), K(+1), Na (+1), Cl(-1), Br(-1), Sr(+2), Ca(+2), O(-2)     .......................1

and

Size

Na+ < Ca2+ < K+ < Sr2+ < Cs+, O2- < Cl- < Br-                ..........................2

so that here

correct answer is A. CsBr

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A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
What would be the vapor pressure at 25.0 °c of a solution of 5.00 g of glucose (c6h12o6) in 100.0 g of ethanol (c2h5oh)?
Crank
Use Raoult's Law:
Psolution = (χsolvent) (P°solvent)

24.90 = (x) (25.756)

x = 0.966765 (this is the solvent mole fraction)

χsolute = 1 - 0.966765 = 0.033235

χsolute = 0.03324 (to four sig figs)
4 0
3 years ago
By definition 12 grams of carbon-12 must contain how many atoms?
krek1111 [17]
It is defined as the amount of a substance that contains as many particles as there are atoms in 12 grams of pure carbon-12. So, 1 molecule contains 6.022×10^23 elementary entities of the substance.

Hope that helps
5 0
3 years ago
As two chlorine atoms combine to form a molecule, energy is
raketka [301]
B. released, forming bonds usually releases energy.
6 0
2 years ago
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