Describe what is happening to the density of the aqueos solution as salt is being added

The water cycle always follows the same pattern

strojnjashka [21]

If it is a true or false questions, the answer is TRUE.

6 0

3 years ago

You and your friend are pushing a bin of books in the same direction. You are pushing with force of 10 N and your friend is push

Mrac [35]

It is 5n , that’s it the difference between the 10n and the 15 n

3 0

3 years ago

Read 2 more answers

Determine the molar concentration of na+ and po4 3- in a 2.25 M Na3 PO4 solution

muminat

Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

B. The concentration of PO4^3- in the solution is 2.25 M.

Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

Na3PO4 will dissociate in solution as follow:

Na3PO4(aq) —> 3Na^+(aq) + PO4^3-(aq)

Thus, from the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+ and 1 mole of PO4^3-

A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

Therefore, the concentration of Na^+ in the solution is 6.75 M

B. Determination of the concentration of PO4^3- in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

7 0

3 years ago

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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3 0

1 year ago

Describe what is happening to the density of the aqueos solution as salt is being added​

Describe what is happening to the density of the aqueos solution as salt is being added