The correct answer that explains similarities between metal and Metalloids as regards the question is They both conduct electricity
- Metalloids can be regarded as elements that are similar to metals, this is because they posses valence orbitals which is described as highly delocalized over macroscopic volumes.
- As a result of this they can serve as electrical conductors.
- metalloids posses small energy gap which is located between the valence band as well as the conduction band, as a result of this they are considered as intrinsic semiconductors when compare to pure conductors like metal.
- Example of metal is Calcium, sodium and that of Metalloids are silicon and germanium
Therefore, metal and Metalloids are similar because of their conductivity of electricity
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Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .
It is given that,
pOH of solution = 7.1
Kw =2.93×10^(-15)
Firstly, we will calculate the value of pKw
The expression which we used to calculate the pKw is,
pKw=-log [Kw]
Now by putting the value of Kw in this expression,
pKw =−log{2.93×10^(-15)}
pKw =15log(2.93)
pKw=14.5
Now we have to calculate the pH of the solution.
As we know that,
pH+pOH=pKw
Now put all the given values in this formula,
pH+7.1=14.5
pH=7.4
Therefore, we find the value of pH of the solution is, 7.4.
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If you mean hydrate as in <em>MgSO4 · 7H2O, </em>then simply find the molar mass of each element you see.
For the example above, that means you would add the molar mass (found on the periodic table) of Mg, then S, then 4(O), 14(H), and 7(O).
The results would be your molar mass for the hydrate.
I hope this is what you meant by your question!