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3 years ago

In which are the ionic solids ranked in order of increasing melting point?

Chemistry

1 answer:

abruzzese [7]3 years ago

7 0

Answer:A

Explanation:

The melting points of solids depend in the relative sizes of ions in the ionic lattice. The smaller the relative sizes of the ions, the higher the lattice energy and the stronger the lattice hence higher melting point. Comparing relative ionic sizes, fluoride ion is lesser in size than chloride ion hence NaF has a higher melting point than NaCl.

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What is the main intermolecular force in H2CO?

aivan3 [116]

Dipole-dipole interactions, and London dispersion interactions

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4 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

The reaction 2A → A2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If

Amiraneli [1.4K]

Answer:

0.19M

Explanation:

2A → A2

Rate constant = 0.0265 M–1min–1

Initial concentration = 2.00 M

Final Concentration = ?

time, t = 180min

The formular relating the parameters is given as;

1 / [A] = kt + 1 / [A]o

1 / [A] = 0.0265 * 180 + (1 / 2)

1 / [A] = 4.77 + 0.5

[A] = 1 / 5.27 = 0.19M

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3 years ago

Which of the following IS TRUE for covalent bonding?

Lina20 [59]

The answer is F : Both A and C

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Explain why it is always necessary to balance a chemical equation. Be sure to discuss atoms, bonds, and the Law of Conservation

Lelechka [254]

All chemical reactions follow the law of conservation of mass. This means that the number of atoms of a specie before the reaction must equal to the number of atoms of that specie after that reaction. The only change that occurs in chemical reactions is the breaking and formation of chemical bonds, due to which energy may be released or absorbed.

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4 years ago

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