Answer:
CO32−
Explanation:
We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.
When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;
2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)
 
        
             
        
        
        
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
                     =  8.6/44
                     = 0.195 moles
now putting the values in equation
V=nRT/P
   = 0.195*0.0821*302/ 0.8762
   = 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
 
        
             
        
        
        
Wym there is no question or statements.
        
             
        
        
        
Answer:
Static Electricity
Explanation:
Most likely, in the room, Jim was building up static electricity by friction with the floor, which he released upon contact with a metal object.
 
        
                    
             
        
        
        
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA] 
 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87