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Gennadij [26K]
3 years ago
8

Given that the Ksp value for Ca3(PO4)2 is 8.6×10−19, if the concentration of Ca2+ in solution is 4.9×10−5 M, the concentration o

f PO3−4 must exceed _____ to generate a precipitate.
Chemistry
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

.0027 M

Explanation:

We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:

Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)

Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2

[PO3−4]=0.0027 M

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Which of the following is a buffer system? Which of the following is a buffer system? H2CO3(aq) and KHCO3(aq) NaCl(aq) and NaOH(
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Answer:

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

In the systems:

H₂CO₃(aq) and KHCO₃(aq): Carbonic acid, H₂CO₃, is a weak acid that, in solution with its conjugate pair, HCO₃⁻ make a <em>buffer system.</em>

NaCl(aq) and NaOH(aq): NaCl is a salt and NaOH is a strong base. Thus, this system <em>is not </em> a buffer system.

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3 0
3 years ago
What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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w of the system is -4578 J and the activity of Sn in the liquid solution  of xsn at 500 degree Celsius is 0.418
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