Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
Sodium reacts with fluorine to make a covalent bond between them. A covalent bond is formed when there is a mutual sharing of electrons between the two adjacent atoms.
Here, Sodium is metal and fluorine is non-metal. So, the bond between them may be an ionic bond, but as the Sodium has one valence electron and fluorine is highly electronegative. Sodium wants to lose electrons and fluorine wants to gain electrons to form a chemical bond. Therefore, the sodium donates its valence electron to fluorine, forms a cation, and fluorine accepts the electron and forms an anion and they distribute the difference in the charge between them.
So, we can say that when sodium and fluorine react to form sodium fluoride with a covalent bond.
Learn more about covalent bonds here:
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Neutrons for example isotopes
The KE can be solve using the formula:
ke = 0.5 mv^2
where m is the mass of the object
v is the velocity
ke = 0.5 ( 2 kg ) ( 5 m/s)^2
ke = 25 J
the GPE can be solved using the formula:
GPE = mgh
where m is the mass
g is the acceleration due to gravity ( 9.81 m/s^2)
h is the height
GPE = ( 4 kg)(9.81 m/s^2)( 40 m)
GPE = 392.4 J
2AgBr+Cl2=2AgCl+Br2
You want everything to be the same on both sides the easiest way to do that is draw a line down the equal sign and write out all you compound and how many there are and do the same on the other side until there even
