Answer:
The sand bath spreads the heat out so that the flask is heated evenly. This reduces the chance of the flask breaking and ensures that there are no hot spots in the reaction mixture which could lead to excessive charring,
Explanation:
The event which is most likely occurring in this scenario is effusion because there is a movement of a gas through a small opening into a larger volume and is denoted as option C.
<h3>What is Effusion?</h3>
This is referred to as the process in which a gas or a substance escapes from a container through a hole of diameter which is usually smaller.
The type of event which is most likely occurring is effusion because of the presence of the small holes in which the balls are made to pass through the center which is why option C was chosen.
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The options are:
- diffusion because particles move from regions of high concentration to regions of low concentration.
- diffusion because particles move from regions of low concentration to regions of high concentration.
- effusion because there is a movement of a gas through a small opening into a larger volume.
- effusion because there is a movement of a gas through a large opening into a smaller volume
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.
The correct answer would be C2H60
Answer:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
