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lisabon 2012
2 months ago
8

A solution is 0.10 M in Pb(NO3)2 and 0.10 M in AgNO3. Solid NaI is added until the second solid compound is on the verge of prec

ipitating.
Which compound precipitates first and what is the I- concentration when the second compound begins to precipitate?Ksp for AgI is 1.5 x 10-16 Ksp for PbI2 is 8.7 x 10-9PbI_2,\:5.9\times10^{-4}\:M
PbI_2,\:4.4\times10^{-3}\:M
AgI,\:2.9\times10^{-4}\:M
AgI,\:9.3\times10^{-5}\:M
AgI,\:8.7\times10^{-8}\:M
Chemistry
1 answer:
Tems11 [2.7K]2 months ago
4 0
The first compound to precipitate will be AgI. Explanation: To form a precipitate from a salt solution, the ionic product must surpass the solubility product. Given that AgI has a significantly low Ksp, it will precipitate before PbI2 does. The concentration of AgI solution also affects precipitation speed; the highest concentration of AgI in the choices provided is...
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In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this
lions [2927]

(a)   Write the balanced half-reactions for the overall process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b)   Assuming E sulfite is 0.57 V, compute E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

3 0
2 months ago
An oxygen atom has a mass of and a glass of water has a mass of . Use this information to answer the question below. Be sure you
VMariaS [2998]

Answer:

Number of moles of oxygen atoms that weigh the same as a glass of water = 3.12 moles

Note: This question lacks certain figures. Below is a complete similar question.

An oxygen atom weighs 2.66*10^-23 g and a glass of water weighs 0.050 kg. What is the weight of one mole of oxygen atoms? Round your result to three significant figures. How many moles of oxygen atoms have a weight equal to the weight of a glass of water? Round your answer to two significant figures.

Explanation:

One mole of a substance comprises the Avogadro number of particles, which is 6.02 * 10²³.

Hence, one mole of oxygen atoms contains 6.02*10²³ atoms.

Weight of a single oxygen atom = 2.66*10⁻²³ g

Weight of one mole of oxygen atoms = weight of a single atom multiplied by the number of atoms in one mole.

Weight of one mole of oxygen atoms = 2.66*10⁻²³ g * 6.02*10²³ = 16.01 g

A glass of water weighs = 0.050 kg or 50 g.

To calculate how many moles of oxygen atoms weigh the same as a glass of water (i.e., 50 g), the following formula is applied;

number of moles = mass/molar mass

mass of oxygen atoms = 50 g, molar mass or weight of one mole of oxygen atoms = 16.01 g

Thus, the number of moles of oxygen atoms = 50 g / 16.01 g = 3.12 moles

4 0
2 months ago
(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
3 months ago
The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?
castortr0y [3046]

Greetings!

The result is:

The new volume is: 2L

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

P_{1}V_{1}=P_{2}V_{2}

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm

Let's put the values into the equation:

2.5L*6.0atm=7.5atm*V_{2}

2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L

Consequently, the new volume is: 2L

Wishing you a lovely day!

7 0
2 months ago
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