I believe the first blank is inertia
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
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 = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Answer:
The new volume of gas would be 30 L.
Explanation:
This is an example of a Combined Gas Laws problem.
Answer:
The answer is 20 % V/V
Explanation:
We use this formula for calculate the %V/V:
%V/V= (ml solute/ml solution) x 100= (75ml/375 ml)x 100 = 20 % V/V
<em>The% V / V represents the amount of ml of solute dissolved in 100 ml of solution</em>
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
