Answer:
44.625862500000004 grams are in 0.376 mol Kbr! Plz mark as brainliest, hope this helped you!
Answer:
Q = 4019.4 J
Explanation:
Given data:
Mass of ice = 20.0 g
Initial temperature = -10°C
Final temperature = 89.0°C
Amount of heat required = ?
Solution:
specific heat capacity of ice is 2.03 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 89.0°C - (-10°C)
ΔT = 99°C
Q = 20.0 g ×2.03 J/g.°C × 99°C
Q = 4019.4 J
B. The reason the temperature experienced no change in group c is because it was likely the control group.
I cannot read question c, the monitor refresh is obscuring the text.
Bronchodilators <span>are prophylactic agents used to treat bronchoconstriction.</span>
Answers-in-bold:
There are two common temperature scales. On the Fahrenheit scale, water freezes at 32 degrees. The Celsius scale divides the interval between the freezing and boiling points of water into 100 degrees.
