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3 years ago

Scientists use what properties to help them identify and classify matter

1 answer:

Volgvan3 years ago

5 0

Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, 
1. Atomic number either atomic mass each element has
2. By substance of that matter either pure substance or mixed substance 
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.

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A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH

Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65

Explanation:

This is the Henderson-Hasselbalch Equation:

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

to calculate the pH of the following Buffer solutions.

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3 years ago

Which of these is an example of a physical change?

trasher [3.6K]

Ain't is the correct answer

8 0

3 years ago

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Where does the solid material go when the solution is made

ELEN [110]

It is dissolved in the salute.

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3 years ago

If the mass of a material is 40 grams and the volume of the material is 9 cm^3, what would the density of the material be?

solniwko [45]

Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?

Density is defined as the ratio of mass and volume.

So,

d = m/V

Using the values, we get

d = 40/9 = 4.44 g/cm³

This means the density of material would be 4.44 g/cm³

3 0

3 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago