Answer:
The identity does not matter because the variables of Boyle's law do not identify the gas.
Explanation:
The ideal gas law confirms that 22.4 L equals 1 mol.
Fluorine 20 (F - Atomic number 9 and atomic mass 20). Firstly we need to know what is beta decay. Beta decay occurs when one neutron changes into a proton and an electron therefore the atomic mass will remain the same as even though we loose a neutron it is replaced by a proton, the atomic number is always raised by 1 when one beta decay occurs. The produced electron is shot out of the nucleus at an incredible speed. This speedy electron we call a beta particle.
Ok now the reaction.
20 20 0
F -> Ne + e
9 10 -1
Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
Hope this helps :).
The answer is (3) methods to achieve racial equality.
Both Du Bois and Washington had different views on how to promote equality, Du Bois emphasized education and immediate political equality while Washington was more concerned with economic equality in industrial and agriculture
<u>Answer:</u> The net ionic equation for the given reaction is 
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:
Ionic form of the above equation follows:
As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation for the given reaction is written above.
Missing in your question :
Ksp of(CaCO3)= 4.5 x 10 -9
Ka1 for (H2CO3) = 4.7 x 10^-7
Ka2 for (H2CO3) = 5.6 x 10 ^-11
1) equation 1 for Ksp = 4.5 x 10^-9
CaCO3(s)→ Ca +2(aq) + CO3-2(aq)
2) equation 2 for Ka1 = 4.7 x 10^-7
H2CO3 + H2O → HCO3- + H3O+
3) equation 3 for Ka2 = 5.6 x 10^-11
HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)
so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)
note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :
CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9
∴ the overall equation will be as we have mentioned before:
when H3O+ = H+
CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55
from the overall equation:
∴K = [Ca2+][HCO3-] / [H+]
when we have [Ca2+] = [HCO3-] so we can assume both = X
∴K = X^2 / [H+]
when we have the PH = 5.6 so we can get [H+]
PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6
so, by substitution on K expression:
∴ 80.55 = X^2 / (2.5 x10^-6)
∴X = 0.0142
∴[Ca2+] = X = 0.0142
