Ans: Calcium sulfate.
K2SO4 (aq) + Ca(NO3)2 (aq) ⇒ 2KNO3 (aq) + CaSO4 (s)
Explanation:
b. What useful functions do oxidation numbers serve?
It is used to show oxidation and reduction (loss and gain of electrons)
b. How many molecules are in 1 mole of molecules?
1 mole = 6.022 * 10^23 molecules
c. What is the name given to the number of molecules in 1 mole?
Avogadro's Number of molecules
21. a. What is the molar mass of an element?
This is the mass of an element divided by the number of moles.
Molar mass = Mass / Number of moles
b. Write the molar mass rounded to two decimal places of carbon, neon, iron and uranium.
amu = Atomic Mass Unit
Carbon = 12.01 amu
Neon = 20.18 amu
Iron = 55.85 amu
Uranium = 238.03 amu
The process through which Polonium is most likely to become stable is: B. alpha decay.
An unstable element refers to a chemical element that lose particles because its nucleus contain an excess of internal energy (neutron or proton).
This ultimately implies that, an unstable element is radioactive in nature.
In Science, some examples of an unstable element are:
Polonium is a chemical element with a large, unstable nucleus.
Basically, the most stable isotope of Polonium is Polonium-209, which typically undergoes an alpha decay to form lead-205 and the emission of an alpha particle.
⇒
----> 
In conclusion, we can deduce from the above chemical equation that Polonium is most likely to become stable through an alpha decay.
Read more: brainly.com/question/18214726
Answer:
The frequency is 
Explanation:
From the question we are told that
The energy required to ionize boron is 
Generally the ionization energy of boron pre atom is mathematically represented as

Here
is the Avogadro's constant with value 
So

=> 
Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

=> 
Here h is the Planks constant with value 
So

=> 
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O