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QveST [7]
3 years ago
6

describe how a physical property such as mass or Texture can change without causing a change in the substance

Chemistry
1 answer:
Tom [10]3 years ago
3 0

You have to be very careful with this question. A change in mass can also occur in chemical changes especially if you have too much of something. For example

CH4 + 1.5 02 ===> CO2 + H2O

If you have too much of either CH4 or O2, there will be some CH4 or O2 left over. There has been a change in mass that you have too much of.

However that is not the point of the question. It is just something you need to be aware of.

Suppose you have a piece of aluminum and you take a course grinder after it. You will change the texture of the side you took the grinder to. If the aluminum has been anodized (a color has been put on it's surface), you may grind the color off or if it is just plain aluminum, you may roughen the surface, but you won't change what the aluminum will do chemically.

You may need only a small portion of the aluminum and you grind off just what you need. That will change the mass of both what you took off and the piece that you want, but the aluminum will still do whatever chemical property you need to use.

So you can change both texture and mass without changing the chemical properties of the substance whose mass or texture you are changing.

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A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
A certain sample of a liquid has a mass of 42 grams and a volume of 35 centimeters3. What is the density of the liquid?
BabaBlast [244]

Answer:

            1.2 g.cm⁻³

Solution:

Data Given:

                             Mass  =  42 g

                             Volume  =  35 cm³

Formula Used:

                             Density  =  Mass ÷ Volume

Putting values,

                             Density  =  42 g ÷ 35 cm³

                             Density  =  1.2 g.cm⁻³

3 0
3 years ago
Read 2 more answers
Describe the term molarity as it relates to an acid or base.
Sidana [21]
Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. Acids and bases can be measured through the concentrations of H+ and OH- ions in units of molarity. Hope this helps.
4 0
4 years ago
Read 2 more answers
Hiw many tons of sulfur are consumed in the production of 98 tons of sulfuric acid?
MariettaO [177]
Well as far as I know to make one ton of sulfuric acid takes 1,000,000 grams, so the answer should be 98,000,000 grams
5 0
4 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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