Answer:
compounds can be broken down to one another
Use the formula for volume. V = lwh. Plug in the numbers for length,
width, and height. Now we have V = 4.0 x 2.0 x 1.0, which is 8.0. So the
volume of the block is 8.0 cm.
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Answer:
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
Explanation:
The two types of acetaldehyde transition are as follows:
n→π* and π→π*
From the attached diagram we have to:
ΔEn→π* < ΔEπ→π*
ΔEα(1/λ)
Thus:
λn→π* > λπ→π*
In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.
The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
Answer:
79.7 mol.
Explanation:
- It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.
<u><em>Using cross multiplication:</em></u>
1.0 mole of tungsten contains → 6.022 x 10²³ atoms.
??? mole of tungsten contains → 4.8 x 10²⁵ atoms.
∴ The no. of moles of tungsten contains (4.8 x 10²⁵ atoms) = (1.0 mol)(4.8 x 10²⁵ atoms)/(6.022 x 10²³ atoms) = 79.7 mol.
Answer:
a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.
b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.
For clear view of the answers: Please refer to calculation 5 in the attachments section.
Explanation:
So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).
Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:
So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.
So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)
Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.
Good Luck Everyone! Hope you will understand.