Answer:
Concentration ammonium is 0.036 g/mL
Explanation:
Stock solution has concentration of 0.108g/mL. Take 10mL stock solution and add 50 mL of water meaning it has 1,08g/60mL. Ammonium sulfate has 2 molecules of cation ammonium cation. So, (1,08g/60mL.)*2=0.036 g/mL
Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)
1) Number of mols of HCl, n
n = 6.3 L * 4.5 mol/L = 28.35 mol
2) ratio: 2 mol HCl / 1 mol Mg = 28.35 mol HCl / x mol Mg
x = 28.35 mol HCL * 1 mol Mg / 2 mol HCL = 14.175 mol Mg
3) Convert mol to mass using atomic mass of Mg
14.175 mol Mg * 24.3 g Mg / mol Mg = 344. 45 g
Answer: 344.45 g
Answer:
2.05mg Fe/ g sample
Explanation:
In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.
In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:
3.465 / 4×100 = <em>86,6%</em>
As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:
1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>
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I hope it helps!
The indirect result of contaminants from commercial farming is algae.
<h3>What is Contaminant?</h3>
Contaminant is a substance either chemical, physical that is very toxic or harmful to living organisms, when it is released into air, water, soil or food.
Contaminant are poisionus substance that are not deliberately or are deliberately introduced to space . This contaminants make things impure and they are highly poisionus.
Therefore, The indirect result of contaminants from commercial farming is algae.
Learn more about contaminant below.
brainly.com/question/465199
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
