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Elis [28]
3 years ago
11

How do we know that caramelizing sugar is a chemical change

Chemistry
2 answers:
galben [10]3 years ago
5 0
Caramelization is a type of browning reaction. As the process occurs, volatile chemicals are released producing the characteristic caramel flavor. The reaction involves the removal of water (steam) and the breakdown of sugar
zavuch27 [327]3 years ago
4 0
The browning reaction causes it to change color. 
You might be interested in
Chemical formula of ammonium oxide
11111nata11111 [884]

The answer is (NH4)2O

8 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)
miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

  • Adding solute to water elevates the boiling point.
  • The elevation in boiling point (ΔTb) can be calculated using the relation:

<em>ΔTb = i.Kb.m,</em>

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

  • van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

<u><em>(1) 1.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

<u><em>(2) 2.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

<u><em>(3) 1.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

<u><em>(4) 2.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

  • <em>So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).</em>

<em></em>

6 0
3 years ago
During an investigation, a student burns magnesium to form magnesium oxide. The starting mass of magnesium is measured as 21.3 g
enot [183]

Answer:

Percentage yield = 85.2%

Explanation:

Given data:

Mass of Mg = 21.3 g

Actual yield of MgO = 30.2 g

Percentage yield = ?

Solution:

Chemical equation:

2Mg + O₂ → 2MgO

Number of moles of Mg = mass/molar mass

Number of moles of Mg = 21.3 g / 24.3 g/mol

Number of moles of Mg = 0.88 mol

Now we will compare the moles of MgO with Mg.

                 Mg           :               MgO

                 2               :               2

                0.88             :             0.88

Mass of MgO:           

Mass of MgO= moles × molar mass

Mass of MgO= 0.88 mol × 40.3g/mol

Mass of MgO =  35.46 g

Actual yield of MgO = 30.2 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield × 100

Percentage yield = 30.2 g/ 35.46 g × 100

Percentage yield = 85.2%

3 0
3 years ago
Fied the emicicol formula of a compoun containing<br> 2.44% hydrogen, 39.02% sulphur, 58.54% oxygen
Viefleur [7K]
The answer is in the attachment below:

7 0
2 years ago
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