A Celsius thermometer with a range of -10°C cannot be used to determine when the soft ball stage is reached in the candy mixture.
Temperature is most commonly measured with a thermometer that has two different scales: Fahrenheit and Celsius. Water freezes at 0°C and boils at 100°C on the Celsius scale and freezes at 32°F and boils at 212°F on the Fahrenheit scale.
Temperature of 236°F corresponds to 113.3°C. Now a Celsius scale has a maximum of 100°C on the thermometer. Hence a Celsius thermometer cannot be used to measure a temperature that corresponds to 236°F.
Answer:
FLUORINE
Explanation:
It has 5 electrons In Its 2P shell!
Answer:
D. Time and pressure changes dead plant material to peat.
Explanation:
There are four steps in the production of coal: peat, lignite, bituminous and anthracite.
Peat: in this stage dead plants are oxidized to water and carbon dioxide and buried under sediments. The partial decomposition of plant matter due to the absence of oxygen is called peat. There is no factor of time and pressure that changes dead plant material to peat.
Lignite: Peat is subjected to heat, pressure, and time to form lignite.
Bituminous: More pressure in lignite removes all the traces of plant matter and form “soft coal”, bituminous coal.
Anthracite: It is the last stage, in which hard coal forms with the combined pressure and high temperature.
Hence, the correct answer is "D. Time and pressure changes dead plant material to peat."
Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
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Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
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W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 205 kPa
= final pressure of gas = ?
= initial volume of gas = 4.0 L
= final volume of gas = 12000 ml = 12 L (1L=1000ml)
(1kPa=0.0098atm)
Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.
