How many liters of water are required to dissolve 1.00 g of barium sulfate?

1 answer:

5 0

<span>It will take 409 liters of water to dissolve 1.00 grams of barium sulfate at 20Â°C Barium sulfate is almost insoluble in water. At 20Â°C, only 2.448 x 10^-3 grams will dissolve in 1 liter of water. So take the amount of barium sulfate and divide by its solubility. 1.00 / 2.448 x 10^-3 = 408.49 liters. Since we only have 3 significant figures, round the result to 3 figures. 409 liters</span>

You might be interested in

The doctor ordered 750ml of a 15 percent solution. the stock solution is 50 percent. how much stock solution and diluent are nee

garik1379 [7]

The doctor required a 15% 750mL solution. To lower the concentration of the solution, you have to add more diluting liquid.

750mL * 0.15 = 112.5mL. This 112.5mL signifies the pure chemical or compound in the solution. This 112.5mL can be taken from the stock solution.

112.5 * 2 = 250mL of stock solution would be needed and 500 mL of the diluting agent should be added. Adding the two together would result to a 750mL solution of 15% concentration.

3 0

2 years ago

Match the term with its correct charge. Positive charge. Negative charge. Neutral: No electric charge. A. electron B. neutron C.

Roman55 [17]

negative charge is electron

positive charge is a proton

no charge is neutron

4 0

3 years ago

Calcium phosphate consists of two ions chemically bonded together. What are charges of each ion?

sdas [7]

Answer: Ca^2+ and P^3-

Explanation:

7 0

3 years ago

What does the 3 indicate in 1s22s22p63s1?

Dahasolnce [82]

The 3 indicates the third electron shell. (Which has only 1 electron in it in this configuration)

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

3 years ago