Answer:
The number of moles = 0.06 moles
Explanation:
i) Formula
Number of moles = Mass of the sample ÷ Molar masses of the substances
ii) Number of moles = 10.0 ÷ (137)+(19×2)
= 10.0 ÷ 175
= 0.06 moles
Answer:
Lonic.an electron will be transferred from potassium to the chlorine atom
Answer : The correct options are,
and 
Explanation :
Single displacement reaction : It is a type of chemical reaction in which the more reactive element displaces the less reactive element.
Option A reaction : 
It is an example of double displacement reaction because in this reaction a positive cation and a negative anion of the two reactants exchange their places to form two new products.
Option B reaction : 
It is an example of single displacement reaction.
Option C reaction : 
It is an example of combination reaction because in this reaction two reactants react to give a single product.
Option D reaction : 
It is an example of decomposition reaction because in this reaction a single reactant decomposes into two or more products.
Option E reaction : 
It is an example of single displacement reaction because in this reaction the most react element, aluminium displaces the less reactive element, hydrogen.
Hence, the options B and E are single displacement reactions.
The answer is D. This is because a synthesis reaction is when two elements combined to make a product (ex. A+B= AB). Hope this helps!
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)