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laiz [17]
4 years ago
6

When a weak base is titrated with a strong acid, the ph at the equivalence point is?

Chemistry
1 answer:
Rainbow [258]4 years ago
6 0
NA depiction of the pH change during a titration of HCl solution into an ammonia solution. The curve depicts the change in pH (on the y-axis) vs. the volume of HCl added in mL (on the x-axis). In strong acid-weak base titrations, the pH at the equivalence point is not 7 but below it.
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Is pepperoni pizza an element, compound,<br> heterogeneous mixture, or homogeneous mixture?
LenKa [72]

Answer: Heterogeneous Mixture

Explanation:

7 0
3 years ago
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Calculate the molarity of a solution prepared by dissolving 11.5 g naoh in enough water to make 1.5 l of solution. 1. 7.6 m 2. 2
zhuklara [117]
Molarity of a solution = (number mol of solute)/ (volume of the solution)

1) Molar mass NaOH= M(Na)+M(O)+M(h)= 23.0+16.0+1.0=40.0 g/mol
2) 11.5 g NaOH *(1mol Naoh/40.0 g NaOH)=0.288 mol NaOH
3)Molarity (NaOH solution) =(0.288 mol)/(1.5 L)=0.192 M≈0.19 M
6 0
4 years ago
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How might a human muscle protein molecule differ from a cow muscle protein molecule?
soldi70 [24.7K]

Human muscle protein differ from a cow muscle protein molecule in the terms of number of mitochondria present, as in human number of mitochondria present is lesser than that of present in cow. Human muscles are not designed to do as much physical work as cows muscles are. Cow's muscles have more mitochondria so that they can generate ATP for physical activity without quickly defaulting to fermentation.  

5 0
3 years ago
Calculate the pH of a 0.020 M H2CO3 solution. At 25 °C, Ka1 = 4.3 × 10-7. H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO3-(aq)
Feliz [49]

Answer:

Explanation:

H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)

Let d be the degree of dissociation

.02( 1-d )                              .02d          .02d

Dissociation constant Ka₁ is given

4.3 x 10⁻⁷  = .02d x .02d / .02( 1-d )

= .004 d² / .02  ( neglecting d in denominator )

= .02 d²

d² = 215 x 10⁻⁷

d = 4.636 x 10⁻³

= .004636

concentration of H₃O⁺

= d x .02

= .004636 x .02

= 9.272 x 10⁻⁵

pH = - log [ H₃O⁺ ]

= - log ( 9.272 x 10⁻⁵ )

5 - log 9.272

= 5 - .967

= 4.033 .

6 0
3 years ago
A student investigates a pure metal, X . The student takes a 100.0 g piece of metal X , heats it to 500.0°C , then places it on
netineya [11]

Answer:

q_metal = -12000 J = -12 KJ

Here, the negative sign indicates that the energy is lost by the metal piece.

Explanation:

The magnitude of energy change of the metal X can be given by the following formula:

q_{metal} = mC\Delta T

where,

m = mass of metal = 100 g

C = Specific Heat Capacity of metal X = 0.24 J/g.°C

ΔT = Change in Temperature of Metal Piece = 0° C - 500°C = -500°C

Therefore, using these values in the equation, we get:

q_{metal} = (100\ g)(0.24\ J/g^oC)(-500^oC)\\<u></u>

<u>q_metal = -12000 J = -12 KJ</u>

<u>Here, the negative sign indicates that the energy is lost by the metal piece.</u>

7 0
3 years ago
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