Answers
Answer:
q= -40.236 or 40.236 KJ of heat released
Explanation:
I attached a picture + a cooling curve.
Related Questions
The blue colour of the sky results from the scattering of sunlight by air molecules. Blue light has a frequency if about 7.5*10^14Hz
Calculate the energy of a mole of photon associated with this frequency
Answers
Answer: The energy of a mole of photon associated with this frequency is [tex]49.5\times 10^{-20}J[/tex]
Explanation:
The energy and frequency are related by :
[tex]E=N\times h\times \nu[/tex]
E = energy of photon
N = number of moles = 1
h = planks constant = [tex]6.6\times 10^{-34}Js[/tex]
[tex]\nu[/tex] = frequency = [tex]7.5\times 10^{14}Hz[/tex]
[tex]E=1\times 6.6\times 10^{-34}Js\times 7.5\times 10^{14}s^{-1}=49.5\times 10^{-20}J[/tex]
The energy of a mole of photon associated with this frequency is [tex]49.5\times 10^{-20}J[/tex]
2. Which of the following is most likely to to cause the extinction
of a species?
A. Mutation
B. A changing environment
C.natural selection
D. Genetic variation
Answers
Answer:
a change of environment
How does the Sun get energy?
Answers
Answer:
Nuclear fushion
Explanation:
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons
Calculate the morality of a 35.4% (by mass) aqueous solution of phosphoric acid (H3PO4). Molar mass of H3PO4 is 98.00g/mol
Answers
Answer:
no idea with the answer pls check with otherr
A 3.6 g sample of iron (III) oxide reacts with sufficient aluminum to be entirely used up.
How much iron is produced?
Answers
Answer:
0.046 mol
Explanation:
Step 1: Write the balanced equation
Fe₂O₃ + 2 Al ⇒ 2 Fe + Al₂O₃
Step 2: Calculate the moles corresponding to 3.6 g of Fe₂O₃
The molar mass of Fe₂O₃ is 159.69 g/mol.
3.6 g × 1 mol/159.69 g = 0.023 mol
Step 3: Calculate the moles of Fe produced from 0.023 moles of Fe₂O₃
The molar ratio of Fe₂O₃ to Fe is 1:2.
0.023 mol Fe₂O₃ × 2 mol Fe/1 mol Fe₂O₃ = 0.046 mol Fe
1. An unknown amount of water
was heated with 3.5 kJ, raising
its temperature from 26°C to
66°C. What was the mass of
the water?
Answers
Answer:
20.93 g
Explanation:
From the question given above, the following data were obtained:
Heat (Q) = 3.5 KJ
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Mass (M) =?
Next, we shall convert 3.5 KJ to J. This can be obtained as follow:
1 KJ = 1000 J
Therefore,
3.5 KJ = 3.5 KJ × 1000 J / 1 KJ
3.5 KJ = 3500 J
Next, we shall determine the change in the temperature of the water. This is illustrated:
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 66 – 26
ΔT = 40 °C
Finally, we shall determine the mass of the water. This can be obtained as follow:
Heat (Q) = 3500 J
Change in temperature (ΔT) = 40 °C
Specific heat capacity (C) = 4.18 J/gºC
Mass (M) =?
Q = MCΔT
3500 = M × 4.18 × 40
3500 = M × 167.2
Divide both side by 167.2
M = 3500 / 167.2
M = 20.93 g
Therefore, the mass of the water is 20.93 g