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BaLLatris [955]
3 years ago
6

Which is something that a PURE SUBSTANCE and a SOLUTION have in common?

Chemistry
1 answer:
zhuklara [117]3 years ago
8 0

D) They both look uniform (the same) throughout.

<h3>Further explanation</h3>

Pure substance can be any element or compound and is formed from one type of atom/molecule only

Meanwhile, the solution is included in a mixture consisting of 2 or more pure substance

Pure substance can be formed through a chemical process while the mixture is through a physical process

Mixture can be separated by physical processes into components of pure substance while pure substance cannot

The mixture itself consists of a homogeneous and heterogeneous solution

The mixture can be divided into a homogeneous mixture if the composition/ratio of each substance in the mixture is the same and a heterogeneous mixture if the ratio of the composition of the substances is not the same (can be varied) in each place.

Mixtures can also be divided into solutions, suspensions, and colloids based mainly on the size of the particles

Homogeneous mixture = Solution

Heterogeneous mixture = suspension, and

The mixture is located between suspension and solution = Colloid

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it is probably acrylic or latex

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H(aq) + NO3 (aq) + HF(aq)

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In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  the major species present are H(aq) + NO3 (aq) + HF(aq).

On the reaction of  HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water  , it will give a polar solution and will form a homogenous mixture.

Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".

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Number of SO2 molecules in 1.28 mol of SO2
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A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v
mihalych1998 [28]

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

    Number of moles  = molarity x volume

Parameters given;

        volume of solution = 500mL  = 0.5L

          molarity of solution = 0.5M

   Number of moles  =  0.5 x 0.5  = 0.25moles

Now to know the volume stock to take;

           Volume of stock = \frac{number of moles }{molarity}

  molarity of stock  = 4M

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3 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
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Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
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