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babunello [35]
3 years ago
9

Aluminum is a metallic element with 3 valence electrons. What is its oxidation number?​

Chemistry
1 answer:
Veronika [31]3 years ago
3 0

Answer:

The neutral atoms of the element have an oxidation number of 0. When the atoms loose their three valence electrons to form ions they then have an oxidation number of +3

Explanation:

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Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
How many orbitals are available in a “d” orbital configuration?
ratelena [41]

Answer:

e.)5

Explanation:

8 0
3 years ago
Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.
N76 [4]

Answer:

1 . Al_2O_3

2. TiO_2

Explanation:

The more stable the ionic compound, the more is it lattice energy.

  • The more the charge on the cation and the anion, the greater is the lattice energy.
  • The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide (Sc_2O_3) is an oxide in which Sc^{3+} behaves as cation and O^{2-} behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . Al_2O_3

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation Al^{3+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

2. TiO_2

This is because the charge on the cation Ti^{4+} is greater than that of Sc^{3+} and also the size of the cation Ti^{4+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

3 0
3 years ago
Is Fe2O3+H2--2Fe+3H2O balanced or unbalanced?
NISA [10]

hydrogen is not balanced

the right balanced equation is

Fe2O3 + 3 H2 -> 2Fe + 3H2O

3 0
3 years ago
The temperature of a sample of water increases from 20c to 46.6c as it absorbs 5650 Joules of heat. What is the mass of the samp
WINSTONCH [101]

Answer: 51 g

Explanation:

5 0
1 year ago
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