Answer:
Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the obstacle, and transmission (accompanied by refraction) into the obstacle or new medium
Answer:
d, 40 dm3.
Explanation:
According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).
From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.
Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).
With that being said, all H2 has reacted, meaning there's no volume of H2 left. 2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)
1 mole of N2 has reacted, meaning from the 30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.
So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.
Answer: True
Air is a mixture of many things; it's not just Oxygen, but Hydrogen, Carbon, and Nitrogen as well are made up of air. Fun fact: Nitrogen is actually more abundant in air
Answer:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
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<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
