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snow_lady [41]
3 years ago
13

Which substance gives heat and light after combustion?

Chemistry
1 answer:
Mariulka [41]3 years ago
7 0

Answer:

Magnesium oxide

Explanation:

The substance which undergoes combustion is said to be combustible.It is called fuel. If a magnesium ribbon is heated,it starts burning. When a magnesium ribbon burns,it combines with the oxygen of air to form magnesium oxide,and liberates heat and light.

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Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
How much thermal energy is in 100 grams of water at 50 degrees Celsius
Ymorist [56]
when the thermal energy is the energy contained within a system that is responsible for its temperature.  

and when the thermal energy  is can be determined by this formula:

q = M * C *ΔT

when q  is the thermal energy

and M is the mass of water = 100 g 

and C is the specific heat capacity of water = 4.18 joules/gram.°C

and T is the difference in Temperature = 50 °C

So by substitution:

∴ q = 100 g * 4.18 J/g.°C * 50

   = 20900 J  = 20.9 KJ
6 0
3 years ago
Read 2 more answers
Which radioactive isotope would take the least amount of time to become stable? rubidium-91 iodine-131 cesium-135 uranium-238
Masteriza [31]
The answer is rubidium -91 because it takes a shorter time of 58.4 seconds to become stable.
5 0
3 years ago
Read 2 more answers
The two different types of color are primary light colors and primary pigment colors. True or false
Nonamiya [84]
Your answer would be True
7 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
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