What is the mass of 0.75 moles of (NH4)3PO4?

Response:

The alteration in color.

Explanation:

The apple browns due to the oxidation process. This occurs when oxygen and water molecules from the air interact with it, resulting in oxidation. The oxidation process is notably rapid at room temperature.

For instance, if the peeled apple is placed in the refrigerator, it takes longer to oxidize and turn brown, but if left at room temperature, it quickly turns brown.

When oxygen interacts with the peeled apple, it activates the polyphenol oxidase enzyme, oxidizing the phenolic compounds and forming quinones, which then react with amino acids to create the brown coloration.

Hello!

To find [H₃O⁺], we will employ the Henderson-Hasselbalch equation, as this involves an acid and its conjugate base:





Next, we derive [H₃O⁺] using the definition of pH:





Hence, the concentration of [H₃O⁺] comes out to be 0.00014 M

Wishing you a good day!

Response:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Clarification:

Balancing a redox reactionis performed using the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform mass balance. This reaction occurs in basic conditions, thus we must add OH⁻ and H₂O as needed.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Ensure electrical balance by incorporating electrons when necessary.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Scale both half-reactions to ensure the electron counts balance out.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Combine both half-reactions and simplify as appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that undergoes reduction is identified as the oxidizer. The species that undergoes oxidation is termed the reducer. In this situation, Br₂ qualifies as both.

MgCO3---> MgO +CO2, x mol MgCO3
M(MgCO3)=24.3+12.0+48.0=84.3 g/mol

CaCO3--->CaO+CO2, y mol CaCO3
M(CaCO3)=40.0+12.0+48.0=100g/mol
M(CO2)=44.0

x*84.3 + y*100=24.00
x*44+y*44=12.00 x+y=12/44=0.2727, x=0.2727-y
(0.2727-y)*84.3 + y*100=24.00
22.99-84.3y+100y=24.00
22.99+15.7y= 24.00, 15.7y=1.01, y=0.06430 mol CaCO3
x=0.2727-0.06430=0.2084 mol MgCO3
0.06430 mol CaCO3*100g/mol=6.43 g CaCO3
0.2084 mol MgCO3*84.3g/mol=17.57 g MgCO3
what does it mean express your answer u?

The question can be easily resolved if we know the abundance of other elements. You would only need to deduct their total from 100. Since this information is missing, the calculation proceeds as follows: Na = 23 g/mol * 1 = 23 g; H = 1 g/mol * 1 = 1 g; C = 12 g/mol * 1 = 12 g; O = 16 g/mol * 3 = 48 g. The total mass then is 84 g. Thus, the percentage of O is calculated as % O = 48/84 * 100 = 57.14%.