Ask question

Ask question

All categories

3 years ago

6

onsider the following equilibrium, in which pure Ag3PO4 is dissolved in liquid water: Ag3PO4(s)↽−−⇀3Ag+(aq)+PO3−4(aq) If the con

centration of the Ag+ ion at a certain point is 6.3×10−5 M, and Ksp=9.1×10−14, will a precipitate form?

1 answer:

telo118 [61]3 years ago

8 0

Answer:

.............................,.....

You might be interested in

is carried out adiabatically in a constant-volume batch reactor. The rate law is Plot and analyze the conversion, temperature, a

Sphinxa [80]

Answer:

See explaination

Explanation:

The mole balance for a constant-volume batch reactor is given such as, For a first-order isothermal reaction, the time to reach a given conversion is the same for constant-pressure and constant-volume reactors. Also, the time is the same for a reaction of any order if there is no change in the number of moles.

Please kindly check attachment for the step by step solution of the given problem.

5 0

2 years ago

Which of these equations is balanced?

kicyunya [14]

I would think that b would be the right answer

7 0

2 years ago

If a block has the mass of 15.00 grams and measurements of 5.00cm by 3.00cm by 2.00cm what is the volume of the block?

Natalija [7]

Answer:

Volume = 30cm³

Explanation:

A block is a geometrical figure and its volume, -look at the figure-, follows the equation:

Volume = Width*Length*Height

As the measurements of the block are 5.00cm, 3.00cm and 2.00cm, the volume is:

Volume = 5.00cm*2.00cm*3.00cm

<h3>Volume = 30cm³</h3>

3 0

2 years ago

The solubility of carbon dioxide in water is 0.161 g CO2in 100 mL of water at 20ºC and 1.00 atmCO2. A soft drink is carbonated w

Sedbober [7]

<u>Answer:</u> The solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of carbon dioxide

$C_2\text{ and }p_2$ are the final concentration and partial pressure of carbon dioxide

We are given:

$C_1=0.161g/100mL\\p_1=1.00atm\\C_2=?\\p_2=5.50atm$

Putting values in above equation, we get:

$\frac{0.161g/100mL}{C_2}=\frac{1.00atm}{5.50atm}\\\\C_2=\frac{0.161g/100mL\times 5.50atm}{1.00atm}=0.886g/100mL$

Hence, the solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

4 0

3 years ago

Given that the formula for mass percent is mass % = mass solutemass solution100%, what will be our starting assumption when conv

attashe74 [19]

Answer:

1. The starting assumption is an amount of 100 grams of solution.

2. It simplifies the problem, because the percentage tells directly the mass of solute present in 100 grams of solution and from that point the calculations are easy to do.

Explanation:

<em>Percentages are always based on the amount of total substance</em>, total solution if it is the case.

If it is mass percentage the amount of total substance is mass.

For instance, a 30% mass solution of sodium chloride means 30 grams of sodium chloride (solute) in 100 grams of solution (includes the solvent plus the solute).

mass percent is mass % = (mass solute / mass solution) × 100%,

<em>Molarity</em> is number of moles of solute per volume in liters of solution:

Molarity = (moles solute / liters solution)

In this case,<em> to convert from percentage to molarity</em> you start assuming a base of 100 grams of solution. For the example of the 30% sodium chloride solution, by assuming 100 grams of solution, you had 30 grams of solute, which you can convert into moles, and, by using the density of the solution (which you need to know), you convert the 100 grams of solution into volume (in liters). Then, you can divide the number of moles of solute into the number of liters of solution, having started with the assumption of 100 grams of solution.

<em>Molality</em> is the number of moles of solute per kilograms of solvent.

Molality = moles solute / kg of solvent.

Again, <em>to convert from percentage to molality</em>, your first step is to assume a base of 100 grams of solution. Then, for the same example of 30% sodium chloride, convert 30 grams of solute to moles, and the 100 grams of solution into 70 grams of solvent (100g solution - 30 grams solute = 70 grams solvent), which is 0.070 kg. Finally, divide the moles of solute by 0.070 kg of solvent.

Then, assuming the amount of solution equal to 100 grams simplifies the problem because the percentage tells directly the amount of solute, which you can convert into moles making calculations easier than with other amount of solution.

5 0

3 years ago