Answer:
The radius of tantalum (Ta) atom is 
Explanation:
From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through
So the volume of the unit cell 
is
We can compute the theoretical density ρ through the following relationship
where
n = number of atoms associated with each unit cell
A = atomic weight
= volume of the unit cell
= Avogadro’s number (
atoms/mol)
From the information given:
A = 180.9 g/mol
ρ = 16.6 g/cm^3
Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.
We can use the theoretical density ρ to find the radio <em>R</em> as follows:
Solving for <em>R</em>
![\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BnA%7D%7B%28%5Cfrac%7B64%5Csqrt%7B3%7DR%5E3%7D%7B9%7D%29N_%7Ba%7D%7D%5C%5C%5Cfrac%7B64%5Csqrt%7B3%7DR%5E3%7D%7B9%7D%3D%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5C%5CR%5E3%3D%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%20%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BnA%7D%7B%5Crho%20N_%7Ba%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%7D)
Substitution for the various parameters into above equation yields
![R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2%5Ccdot%20180.9%7D%7B16.6%5Ccdot%206.023%20%5Ctimes%2010%5E%7B23%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cfrac%7B64%5Csqrt%7B3%7D%7D%7B9%7D%7D%7D%5C%5CR%20%3D%201.43%20%5Ctimes%2010%5E%7B-8%7D%20%5C%3Acm%20%3D%200.143%20%5C%3Anm)
Answer:
B
Explanation:
Please take a look at the picture attached for the drawings and structures.
C2H4 is a alkene (C-C double bond). When steam (water) is added, it turns into an alcohol, where the double bond breaks and a (-OH) functional group is attached to one of the Carbons. In this case, C2H4 ethene is turned into ethanol.
When an alcohol undergoes oxidation, primary alcohols turn into aldehyde (-CHO) or carboxylic acids (-COOH). Secondary alcohol turns into ketone. Ethanol is a primary alcohol. And since it later reacts with propanol, it can only form carboxylic acid when it oxidizes. The product in this reaction is ethanoic acid.
Carboxylic acid reacts with alcohol to form an ester (-COO-). the -COOH group from acid combines with the -OH group from alcohol to form an ester bond -COO- while eliminating H2O. Therefore, when propanol undergoes esterification with ethanoic acid, propyl ethanoate is produced. It is the answer of B.
Your nuclear reactor rots
Answer:
Explanation:
C₆H₅NH₂ + H₂O = C₆H₅NH₃⁺ + OH⁻
C₆H₅NH₃⁺ is also called anilinium ion . Thus we see that aniline reacts with water to form anilinium ion and forms OH⁻ ion . Due to it, it shows basic properties . The pH value of this aqueous solution is more than 7 . But it is a weak base because the concentration of OH⁻ formed is very less . It is so due to its less value of Kb . It has been explained as follows
Kb = ![\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2][H_2O]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC_6H_5NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BC_6H_5NH_2%5D%5BH_2O%5D%7D)
Since Kb is 4.3 x 10⁻¹⁰ which is very low value so concentration of OH⁻ that is [OH⁻] formed will also be very low . Hence it will be a weak base.
