Answer:
(a) H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Explanation:
The equation for your buffer equilibrium is:
H₃PO₄(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq)+ H₂PO₄⁻(aq)
(a) Adding H₃O⁺
The hydronium ions react with the basic dihydrogen phosphate ions.
H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) Adding OH⁻
The OH⁻ ions react with the more acidic hydronium ions.
OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Answer:
Follow these steps.
1. Fill the matchbox with pebbles. Weigh the matchbox with the pebbles inside. Record that weight.
2. Tie the string to the box. Allow the string to hang over the edge of the table.
3. Tie the other end of the string to a corner of the plastic bag, leaving an opening to put in coins.
4. Add coins one by one until the box is pulled off the table.
5. Count and record the number of coins and the weight of the bag with the coins in it.
6. Lay the round sticks on the table about 1 inch apart and about 2 inches from the edge of the table.
7. Put the matchbox on the rollers farthest from the edge of the table.
8. Now add coins one by one to the bag until the box is pulled off the table.
9. Count and record the number of coins and the weight of the bag with the coins in it.
10. Repeat the experiment. Determine your margin of error if your results vary. For accuracy, repeat the experiment if desired.
11. Using the equation for the coefficient of friction in the text above, determine the coefficient of friction for the matchbox in each experiment. Include this data in your summary.
Explanation:
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Answer:
- 278.85 J
Explanation:
Given that:
Pressure = 1.1 atm
The initial volume V₁ = 0.0 L
The final volume V₂ = 2.5 L
The work that takes place in a reaction at constant pressure can be expressed by using the equation:
W = P(V₂ - V₁ )
Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:
W = - P(V₂ - V₁ )
W = -1.1 atm ( 2.5 - 0.0) L
W = -1.1 atm (2.5 L)
W = -2.75 atm L
Recall that:
1 atm L = 101.4 J
Therefore;
-2.75 atm L = ( -2.75 × 101.4 )J
= -278.85 J
Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J
Protons= 12, Electrons= 12, neutrons= 12