Answer:
THE NEW PRESSURE OF THE HELIUM GAS IS 124kPa AFTER THE VOLUME WAS INCREASED FROM 2.48 L TO 2.98 L
Explanation:
Using Boyle's law which states that at constant temperature, the pressure of a given gas is inversely proportional to the volume occupied by the gas.
Mathematically,
P1 V1 = P2 V2
P1 = 150 kPa = 150 * 10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the formula making P2 the subject of the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 * 10^3 * 2.48 / 2.98
P2 = 372 * 10 ^3 / 2.98
P2 = 124.83 * 10^3 Pa or 124.8kPa
In other words, the new pressure of the helium gas after its volume was increased from 2.48 L to 2.98 L is 124.8kPa.
Answer:
the correct answer is option C. Na
20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal; = 20.6kCal = 20600Cal
learn more:
Specific heat brainly.com/question/3032746
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The answer is 14.22 mg / (mm^2)