Balancing redox reactions:
Oxygen should be balanced by adding 
as needed, while hydrogen should be balanced by adding 
.
What is a redox reaction?
Redox reactions, also known as oxidation-reduction reactions, involve the simultaneous oxidation and reduction of two different reactants.
The Half-Equation Method is one technique used to balance redox processes. The equation is divided into two half-equations using this technique: one for oxidation and one for reduction.
By changing the coefficients and adding , , and 
in that order, each reaction is brought into equilibrium:
- By putting the right number of water () molecules on the other side of the equation, the oxygen atoms are brought into balance.
- By adding ions to the opposing side of the equation, one can balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom).
- Total the fees for each side. Add enough electrons () to the more positive side to make them equal. (As a general rule, and are nearly always on the same side.)
- The on either side must be made equal; if not, they must be multiplied by the lowest common multiple (LCM) in order to make them equal.
- One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
- Now that the equation has been verified, it can be balanced.
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It is false. The salt bridge is not a path for electrons, but a path for ions to flow from one half-cell to another. It help to balance the charge between the oxidation and reduction vessels.
Answer:
29.575%
Explanation:
Data provided:
Calories taken in daily diet = 2000
Recommended amount of fat = 65 grams
Average number of calories for fat = 9.1 calories / g
Thus,
Number of calories in the diet with average number of calories for fat
= Recommended amount of fat × Average number of calories for fat
= 65 × 9.1
= 591.5 calories
Therefore,
the percentage of calories in his diet supplied = ( 591.5 / 2000 ) × 100
= 29.575%
Answer:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
We can use the ideal gas law equation to find the pressure
PV = nRTwhere
P - pressure
V - volume - 2.6 x 10⁻³ m³
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³ = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm
