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3 years ago

What is the equation for the base ionization constant of PO4^3-

1 answer:

3 0

Chemical reaction of PO₄³⁻ ion in water:
PO₄³⁻(aq) + H₂O(l) → HPO₄²⁻(aq) +OH⁻(aq).
Kb = [HPO₄²⁻] · [OH⁻] / [PO₄³⁻]; <span>base ionization constant.
</span>Base ionization constant <span>is the equilibrium </span>constant<span> for the </span>ionization<span> of a </span>base<span>.
</span>According to Bronsted-Lowry theory acid are donor of protons and bases are acceptors of protons (the hydrogen cation or H⁺<span>).
</span>PO₄³⁻ is Bronsted base and it can accept proton and become conjugate acid HPO₄²⁻.

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Question 3 (5 points)

Daniel [21]

Answer:

332.918g O2

Explanation:

I'm having some issues with the work however, your final answer should be 332.918g O2

Hope this helped!

8 0

2 years ago

Unit 1- Unit Assessment (Matter & Density)

konstantin123 [22]

Answer:

<h3>The answer is 11 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 3025 g

volume = 275 mL

We have

$density = \frac{3025}{275} \\$

We have the final answer as

Hope this helps you

7 0

2 years ago

The osmotic pressure of a solution formed by dissolving 35.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25°c is __________ at

Lunna [17]

Mass of aspirin = 0.025 g
Molar mass of C9H8O4 is 180.1583 g/mol
moles of aspirin = .025g / 180.1583 g/mol = 0.000138767 moles
volume solution = .250 L
molarity of the solution = 0.000138767 moles / .250L =5.551 x 10 ^-04 Moles / liter
for aspirin i = Vant'Hoff factor = 1 particle in solution
T = 25 + 273 =298 K
osmotic pressure = M x R x T x i =
5.551 x 10 ^-04 mole L -1 x 0.08206 L atm K−1 mol−1 x 298 K x 1 = 0.0136 atmospheres

6 0

2 years ago

Read 2 more answers

Give two examples of halohydrin formation

BigorU [14]

Answer:

Bromohydrin and chlorohydrin are examples of halohydrins (where X = Br or Cl).

8 0

1 year ago

Which of the following would you except to see in the death of a star that is less than 0.5 solar mass

Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: $1.4 \;M_\odot$ .

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its $0.5 \;M_\odot$ , which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0

2 years ago