I'll see what I can do here...
1) Nonmetal
2) Calcium (Ca), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table.
3) Hafnium
4) 204.3833 u
5) Not sure what you're asking, but oble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)
6) The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At)
7) The Actinide series contains elements with atomic numbers 89 to 103 and is the third group in the periodic table.
8) 33
9) 88
10) 30
Hope this helps!
V ( HCl ) = 16.4 mL / 1000 => 0.0164 L
M( HCl) = ?
V( KOH) = 12.7 mL / 1000 => 0.0127 L
M(KOH) = 0.620 M
Number of moles KOH:
n = M x V
n = 0.620 x 0.0127
n = 0.007874 moles of KOH
number of moles HCl :
<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
</span>
moles HCl = 0.007874 * 1 / 1
= 0.007874 moles of HCl
M = n / V
M = 0.007874 / <span>0.0164
</span>= 0.480 M
Answer (2)
hope this helps!
We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
Explanation:
because there are 4 Iodines on the left, we'll put. 4 in front of NaI to balance it. This would result in 4 Na on the left, so we'll put a 2 in front of Sodium Sulfate to balance the right side. Now we have 4 Na and I on both side, as well as 2 Sulfate on both sides. Pb is already balanced. The equation is now complete.
