Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)
Answers
The balanced redox reaction is
5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)
The unbalanced redox reaction is :
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
The above equation can be balanced by ensuring the atom of the elements
on the left hand side is equal to those on the right hand side.
We have 3 atoms of Mn on the left side of the equation and 1 atom on the
right hand side.This will be balanced by putting 3 in front of MnO₄− as shown
below:
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 3 atoms of O on the left hand side and 12 atoms of O on the right
hand side. This is balanced by putting 5 in front of AsO₂− and 2 in front of
H₂O as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 4 atoms of H on the left hand side and 1 atom of H on the right
hand side.We can balance by putting 4 in front of H⁺ as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)
We have 5 atoms of As on the left hand side and 1 atom of As on the right
hand side. We can balance by putting 5 in front of As as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)
The equation is therefore now balanced as the number of atoms of the
element on the left hand side are equal with those on the right hand side.
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Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
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You are given a glass rod that is negatively charged and a pinwheel that has a negative charge. What will happen when the glass rod is brought close to the pinwheel? Why?
Answers
Answer:
THE NEW FREEZING POINT IS -4.196 °C
Explanation:
ΔTf = 1 Kf m
molarity of MgCl2:
Molar mass = (24 + 35.5 *2) g/mol
molar mass = 95 g/mol
7.15 g of MgCl2 in 100 g of water
7.15 g = 100 g
(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3
= 0.715 g /dm3
Molarity in mol/dm3 = molarity in g/dm3 / molar mass
= 0.715 g /dm3 / 95 g/mol
m = 0.00752 mol/ dm3
So therefore:
ΔTf = i Kf m
1 = 3 (1 Mg and 2 Cl)
Kf = 1.86 °C/m
M = 0.752 moles
So we have:
ΔTf = 3 * 1.86 * 0.752
ΔTf = 4.196 °C
The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C
Answers
Answer:
solubility of X in water at 17.0 is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = = 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
Answers
Answer: The final temperature of the sample is 62.66°C
Explanation:
To calculate the amount of heat absorbed, we use the equation:
where,
Q = heat absorbed = 16.7 kJ = 16700 J (Conversion factor: 1 kJ = 1000 J)
m = Mass of the sample = 225 g
c = specific heat capacity of sample =
= change in temperature =
Putting values in above equation, we get:
Hence, the final temperature of the sample is 62.66°C
Answers
Answer:
p orbitals only
Explanation:
Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.
Final answer:
Carbon's valence electrons reside in the 2s and 2p orbitals. These orbitals hybridize during bond formation to create equivalent sp3 hybrid orbitals, as evidenced in the methane molecule. Carbon's valence electrons are not placed in d orbitals.
Explanation:
Carbon (atomic number 6) has a total of six electrons. Two of these fill the 1s orbital. The next two fill the 2s orbital, and the final two are in the 2p subshell. According to Hund's rule, the most stable configuration for an atom is one with the maximum number of unpaired electrons. Therefore, carbon has two electrons in the 2s subshell and two unpaired electrons in two separate 2p orbitals. When discussing valence electrons, the electrons in the outermost shell are the ones considered, which for carbon are the electrons in the second shell namely 2s and 2p.
The geometry of the methane molecule (CH4) illustrates that in the bonding process, the s and p orbitals hybridize to allow the formation of four equivalent bonds with hydrogen atoms. Without hybridization, we would expect three bonds at right angles (from the p orbitals) and one at a different angle (from the s orbital). Nonetheless, through orbital hybridization, all four bonds in methane are identical, which is explained by the concept of sp3 hybridized orbitals.
Therefore, the valence electrons for carbon would be placed in the s orbital and p orbitals, not in the d orbitals, because carbon does not have electrons in the d subshell in its ground state. Additionally, the s and p orbitals are the only ones involved in bonding for carbon in most of its compounds, such as methane.
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Answers
Answer:
enantiomeric excess = 68%
Explanation:
Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:
ee = R - S / R + S * 100
Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.
Replacing with values of the problem:
ee = 84% - 16% / 84% + 16% * 100
ee = 68%
Final answer:
The enantiomeric excess of the mixture, defined as the difference between the concentrations of the R and S enantiomers, is 68.0%.
Explanation:
The enantiomeric excess (ee) is defined as the absolute difference between the mole percentage of the major enantiomer and the minor enantiomer in a mixture. In a sample that contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, the enantiomeric excess is calculated as follows:
- Calculation: The enantiomeric excess is 84.0% (R) - 16.0% (S) = 68.0%
Therefore, the enantiomeric excess of the mixture is 68.0%.
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Answers
There is 65% of NaHCO3 in the sample.
The equation of the reaction is;
HA + NaHCO3 -----> NaA + CO2 + H2O
Amount of CO2 formed = mass/molar mass
mass of CO2 = 0.561 g/44 g/mol = 0.013 moles
From the balanced reaction equation;
1 mole of NaHCO3 yields 1 mole of CO2
0.013 moles of Na2CO3 yields 0.013 moles of CO2
Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3
Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1
= 65%
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Answer:
63.75%.
Explanation:
The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;
HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).
Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.
Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.
Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.
Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.
Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.