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Cloud [144]
3 years ago
5

Assume you have 4 solids (A, B, C and D) of similar mass. Which of these requires the greatest energy input to melt?

Chemistry
1 answer:
bazaltina [42]3 years ago
7 0

The solid that require the greatest energy input to melt by mass is the option;

Covalent network

Solution Explanation

The elementary particles of a solid are held together by bonds that require an input of energy to unlock, and once broken, the particles are then able to change location within their containing vessels with less restrictions

Polar covalent molecular solids have the following characteristics;

a) Soluble in water b) Low melting point, b) Conduct electricity

Solids that are made up of a covalent network have the following characteristics

a) High melting point temperature b) Non conductive of electricity c) Not soluble in water

Solids of ionic compounds have the following characteristics;

a) High melting point temperature b) The liquid state and solution conducts electricity c) Soluble in water

Solids that have nonpolar covalent bonds have;

a) Low melting point b) Normally in the gaseous or liquid state b) Not water soluble

The covalent network, and the solids ionic compounds require the most energy to melt, however, the strength of the ionic bond in an ionic compound is a factor the charges present and the sizes of the atom, while the covalent network solid, are combined to <em>form essentially as a single molecule</em> and therefore require the greatest heat energy input to melt

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If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
Vanillin is the compound containing carbon, hydrogen, and oxygen that gives vanilla beans their distinctive flavor. The combusti
nignag [31]
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C 14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O molar mass of C=12 g/mole molar mass of H=1 g/mole 0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C 0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O molar mass of O=16 g/mole 9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602 C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602 C2.66H2.66O1 is the empirical formula; to obtain whole numbers multiply by 3 3[C2.66H2.66O1] = C8H8O3 above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu The empirical formula weight and the molecular formula weight are the same . Molecular formula is C8H8O3.</span>
7 0
3 years ago
An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4. What is the concentration of hydronium ion in
Alex_Xolod [135]

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

Methylamine is an amine which is an organic weak base. Its chemical formula is CH₃NH₂. When it undergoes hydrolysis wherein water is acting as an acid, the reaction would be

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Initial                    0.05          -                 0         0

Change                 -x                              +x       +x

----------------------------------------------------------------------------

Equilibrium         0.05-x                           x          x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

#SPJ4

5 0
2 years ago
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

3 0
3 years ago
When the n quantum number equals 1, we are in what orbital?
g100num [7]

Answer:

p orbital

Explanation:

8 0
3 years ago
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