I'm having trouble with this one. At the very least, they both purify water. If I'm not mistaken, distillation is part of the process of purifying water.
Really sorry, I hope I helped at least a little! If you have any other questions I might be able to answer better, let me know..
The reaction between ammonium sulfate and calcium hydroxide is given below.
(NH₄)₂SO₄ + Ca(OH)₂ --> 2NH₃ + CaSO₄ + 2H₂O
From the balance equation, we can conclude that every 74 g of calcium sulfate reacted with enough amount of ammonium sulfate will yield 34 grams of ammonia. From the given amount,
(20 g calcium sulfate) x (34 grams ammonia / 74 g calcium sulfate)
= <em>9.19 g ammonia</em>
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
Learn more about heat transfer:
brainly.com/question/10286596
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The appearance of the protist differ from that of the onion sample due to the presence of motile structures.
<h3>What is a Protist?</h3>
This is an eukaryote which could be unicellular or multicellular and examples include protozoa etc.
The major difference between a protist and onion sample is that protists have motile structures such as flagella, cilia etc while plant cells such as onions don't have.
Read more about Protist here brainly.com/question/2169979
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Temp must be Kelvin
38 C =
<span>
<span>
<span>
311.15
</span>
</span>
</span>
K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
<span>
<span>
<span>
7.1182746306
</span>
</span>
</span>
Liters
The formula to use is:
Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)
