The characteristics of wave are amplitude, speed , frequency and wavelength.
<u>Explanation:</u>
Amplitude is defined as the distance covered by the particles or molecules moving in a wave nature. So the maximum positive and negative peak position is termed as amplitude of the wave.
Every wave forms a series of crests and trough regions. The crest region is the positive amplitude while the trough region is the negative amplitude. The distance between two successive crests and troughs are termed as wavelength.
A single crest and trough forms a complete cycle. So the number of times a cycle is repeated for a given time is known as frequency. Speed is defined as the product of frequency and wavelength of the wave. In other words, speed of a wave is the rate at which the particles are moving a given area per unit time.
Answer:
1.4 × 10² mL
Explanation:
There is some info missing. I looked at the question online.
<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>
Step 1: Given data
- Initial volume (V₁): 215 mL
- Initial pressure (P₁): 625 mmHg
- Final pressure (P₂): 1.3 atm
Step 2: Convert 625 mmHg to atm
We will use the conversion factor 1 atm = 760 mmHg.
625 mmHg × 1 atm/760 mmHg = 0.822 atm
Step 3: Calculate the final volume of the air
Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL
Answer:
To monitor Pressure increase
Explanation:
the more pressure, the higher the production of the gas, which can be recorded, more pressure meaning higher production. Then vary the temperature.
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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Mn₂O
Explanation:
The oxide that will most likely form colored solutions is Mn₂O.
This is because most transition metals form colored compounds. Manganese is a transition metal belonging to the d-block on the periodic table.
- Other examples of transition metals are scandium, titanium, iron, copper, cobalt, nickel, zinc
- They belong to the d-block on the periodic table.
- They have variable oxidation states.
- Most of their solutions are always colored.
Learn more:
Periodic table brainly.com/question/8543126
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