What atomic component contributes to many of the trends of elements

For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Ag + is reduced to Ag . Complete and balance the equation for t

Sunny_sXe [5.5K]

Answer:

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺ → Ag

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ → Ag (1)

Now, we need to balance the half-reaction of Cr:

Cr → CrO₄²⁻

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:

Cr + OH⁻ → CrO₄²⁻ + 6e⁻

The hydrogen atoms will be balanced using H₂O molecules:

Cr + OH⁻ → CrO₄²⁻ + 6e⁻ + H₂O

The balanced equation is:

Cr + 8OH⁻ → CrO₄²⁻ + 6e⁻ + 4H₂O (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ → Ag)

Cr + 8OH⁻ → CrO₄²⁻ + 6e⁻ + 4H₂O

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

7 0

4 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

3 years ago

What is the concentration of a solution that has a volume of 2.5L and contains exactly 2.125 moles of calcium phosphate

Iteru [2.4K]

Answer:

0.85 mol/L.

Explanation:

Molarity is defined as the no. of moles of solute dissolved in a liter of the solution.

M = (no. of moles of solute)/(Volume of the solution (L))

no. of moles of calcium phosphate = 2.125 mol.

Volume of the solution = 2.5 L.

∴ M of calcium phosphate = (2.125 mol)/(2.5 L) = 0.85 mol/L.

7 0

3 years ago

A 4.5 M solution is to be diluted to 750.0 mL of a 1.5 M solution. How many mL of the 4.500 M solution are required?

maw [93]

Answer:

309

Explanation:

7 0

3 years ago

What is the compound formula for Ba2 and NO3–?

Dimas [21]

That would be the compound Ba(NO3)2. The Barium ion has a charge of 2+ so there are 2 NO3- ions in the formula to produce a neutral compound.

3 0

3 years ago

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