All categories

2 years ago

(1) The solubility of Salt AB2(S) IS 5mol/dm^3.

Chemistry

1 answer:

Svetlanka [38]2 years ago

6 0

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s] [s] [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

You might be interested in

CO2 is an element. True False

Reil [10]

Answer:False

Explanation:co2 or carbon dioxide is a compound not an element, you can even check this on the periodic table

7 0

3 years ago

Read 2 more answers

Which cell organelle is like the construction site, where proteins are built? Which cell organelle packages and transports the p

german

Explanation:

The endoplasmic reticulum consists of a network of a tube-like passageway through which proteins from the ribosomes are able to be moved within a cell as the road system allows for movement throughout the city.

7 0

3 years ago

Fill in the coefficients that will balance the following reaction a0h2so4 + a1koh → a2k2so4 + a3hoh

Anuta_ua [19.1K]

A0 = 1
a1 = 2
a2 = 1
a3 = 2

This can be solved by guessing and checking and making sure that all atoms on the left side are accounted for on the right side. Both sides must have the same amount of atoms or the balancing is not correct.

7 0

3 years ago

Read 2 more answers

40 POINTS PUT A GOOD ANSWER

Andreyy89

Answer:

B What role do animals have in the carbon cycle?

Group of answer choices

use water and release oxygen

use sugar and release carbon

use carbon and release oxygen

use carbon and release nitrogen

D, E Select TWO statements that describe what the scientists will observe in the ecosystem after fossil fuels have been extracted and burned.

Group of answer choices

There is more available space for carbon to become fossil fuels.

There is less carbon in storage.

There will be less carbon in the water.

There is more carbon in the atmosphere.

There will be more carbon respired by organisms.

There will be less carbon released through decomposition.

Explanation:

Animals obtain energy through sugar from the food they eat and release carbon into the air through respiration. Therefore, they partake in the carbon cycle by using sugar and releasing energy.

As far as the second question, when fossil fuels are burned, the carbon contained within them is released into the atmosphere in the form of carbon dioxide. So, there will obviously be a greater amount in the atmosphere.

I would like to apologise in advance if either of my option choices were incorrect.

6 0

2 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

2 years ago