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Rom4ik [11]
4 years ago
6

Give five differences between physical and chemical change​

Chemistry
2 answers:
Nostrana [21]4 years ago
4 0
Remember when it’s chemical substances change dramatically
aleksley [76]4 years ago
3 0

Answer:

When a substance undergoes a physical change, its composition remains the same despite its molecules being rearranged. When a substance undergoes a chemical change, its molecular composition is changed entirely. Thus, chemical changes involve the formation of new substances.

Physical change is a temporary change. A chemical change is a permanent change.

A Physical change affects only physical properties i.e. shape, size, etc. Chemical change both physical and chemical properties of the substance including its composition

A physical change involves very little to no absorption of energy. During a chemical reaction, absorption and evolution of energy take place.

Some examples of physical change are freezing of water, melting of wax, boiling of water, etc. A few examples of chemical change are digestion of food, burning of coal, rusting, etc.

Generally, physical changes do not involve the production of energy. Chemical changes usually involve the production of energy (which can be in the form of heat, light, sound, etc.)

In a physical change, no new substance is formed. A chemical change is always accompanied by one or more new substance(s).

Physical change is easily reversible i.e original substance can be recovered. Chemical changes are irreversible i.e. original substance cannot be recovered.

Explanation:

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PLEASE HELP!! WILL MARK BRAINLIEST AND THANK YOU!!!
Pepsi [2]

Answer:8

Everything after the decimal place is a significant figure here

5 0
4 years ago
What problem would a person have if the nucleic acid in one of his or her
Dmitry_Shevchenko [17]
The answer is B i think
7 0
3 years ago
Suppose that you have a 60.0% solution of NaOH. How many milliliters of water must be added to 30.0 mL of this solution to prepa
defon

Answer:

  • <u>21.4 ml (second choice)</u>

Explanation:

<u>1) Data:</u>

a) C₁ = 60.0% (initial solution)

b) V₁ = 30.0 ml (initial solution)

c) C₂ = 0% (pure water)

d) V₂ = ? (pure water)

e) C₃ = 35.0% (final concentration)

<u>2) Formula:</u>

  • C₁V₁ + C₂V₂ = C₃V₃
  • V₁ + V₂ = V₃ (assuming volume addtivity)

<u>3) Solution:</u>

<u />

a) Substitute values in the first formula:

  • 60.0% × 30 ml + 0 = 35% (30 ml + V₂)

b) Solve the equation (units are supressed just to manipulate the terms)

  • 18 = 10.5 + 0.35V₂

  • 0.35V₂ = 18 - 10.5 = 7.5

  • V₂ = 7.5 / 0.35 = 21.4 ml ← answer    
3 0
4 years ago
A 1.900 g sample of C6H12 is burned in an excess of oxygen. What mass of C)2 and H20 should be obtained
Alexeev081 [22]
The balanced chemical reaction is:

<span>C6H12 + 9O2 = 6CO2 + 6H2O

We are given the amount of </span><span>C6H12 to be used for the reaction. THis will be the starting point for the calculations.

1.900 g </span>C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol CO2 / 1 mol <span>C6H12) (44.01 g /mol ) = 5.96 g CO2
</span>1.900 g C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol H2O / 1 mol <span>C6H12) (18.02 g /mol ) =  2.44 g H2O
</span>
6 0
3 years ago
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
Vinvika [58]

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.

Answer:

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is as follows:

C_2H_5NH_{2(aq)     +     H^+_{(aq)      ⇄        C_2H_5NH_{3(aq)}^+

Given that concentration of C_2H_5NH_{2(aq) = 10%

i.e 10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = \frac{10}{45.08}

= 0.222 mol

Molarity of C_2H_5NH_{2(aq) = 0.222 × \frac{1000}{100}mL

= 2.22 M

However, number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

number of moles of C_2H_5NH_{2(aq) = 20 mL × 2.22 M

= 44*10^{-3} mole

Concentration of C_2H_5NH_{2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq)  is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

Finally, the pH value at equivalence point is:

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

∴

The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.

6 0
4 years ago
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