If the uncertainty of a certain measurement instrument is not given, then it is assumed to be equal to half of the least count of that instrument. In this case, the least count is 10 ml, so half of this is 5 ml. Therefore, the graduated cylinder has an uncertainty of +/- 5 ml
Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
I looked up what is the molecular formula for Phosphine and got this: PH3
Hope this helps! Let my know if this was correct.
You can answer this question by only searching the element in the periodic table.
The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).
The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
Answer:
0.3793 M
Explanation:
The unknown metal is zinc. So the equation of the reaction is;
Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)
From Nernst equation;
E = E° - 0.0592/n log Q
[Cu2+] = 0.050179 M
n = 2
[Zn^2+] = ?
E = 1.074 V
E° = 0.34 - (-0.76) = 1.1 V
Substituting values;
1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179
1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179
-0.026 = -0.0296 log [Zn^2+]/0.050179
-0.026/-0.0296 = log [Zn^2+]/0.050179
0.8784 =log [Zn^2+]/0.050179
Antilog(0.8784) = [Zn^2+]/0.050179
7.558 = [Zn^2+]/0.050179
[Zn^2+] = 7.558 * 0.050179
[Zn^2+] = 0.3793 M
