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3 years ago

12

What is one way to test whether an unknown solution is acidic or basic

1 answer:

Dima020 [189]3 years ago

4 0

A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale. For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.

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4 years ago

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Where are the alkali metals and alkaline earth metals and halogens and noble gases located in the periodic table?

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3 years ago

What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 321 m

xxMikexx [17]

Answer: The osmotic pressure of a solution is 53.05 atm

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (methanol) = 22.3 g

Volume of solution = 321 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$\pi=1\times \frac{22.3\times 1000}{32.04\times 321}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K$

$\pi=53.05atm$

Hence, the osmotic pressure of a solution is 53.05 atm

7 0

3 years ago

A chemist adds 1.80L of a 1.1/molL aluminum chloride AlCl3 solution to a reaction flask. Calculate the millimoles of aluminum ch

amm1812

Answer:

2000 millimoles of AlCl₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 1.8 L

Molarity of solution = 1.1 mol /L

Millmole of AlCl₃ =?

Next, we shall determine the number of mole of AlCl₃ in the solution.

This can be obtained as follow:

Volume of solution = 1.8 L

Molarity of AlCl₃ solution = 1.1 mol /L

Number of mole of AlCl₃ =?

Molarity = mole /Volume

1.1 = Number of mole of AlCl₃ / 1.8

Cross multiply

Number of mole of AlCl₃ = 1.1 × 1.8

Number of mole of AlCl₃ = 1.98 moles

Finally, we shall convert 1.98 moles to millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

1.98 mole = 1.98 mole × 1000 millimoles / 1 mole

1.98 mole = 1980 millimoles

1.98 mole ≈ 2000 millimoles

Thus, the chemist added 2000 millimoles of AlCl₃

7 0

3 years ago

Rectangular prism has a height of 3cm, a width of 7 cm, and a length of 5 cm. What is it's volume?

goldfiish [28.3K]

Volume= length•width•height

V=5•7•3

V= 105cm^3

6 0

3 years ago