Temperature of 62.0 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium
Answer:
There must be an equal amount of each element on both sides of the equation.
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Answer:
5746.0 mL.
Explanation:
We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.
V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>
Soil covered, saturated, submerged, flooded w water, standing water
A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4)
0 = -25t+50+19.6t - 19.6
5.4t = 30.4
t = 5.62962963 s
b) h = -4.9(5.62962963)^2
h = -155.2943759
the building is 155.2943759 m high
c) speed 0of first stone
= at
= 9.8*5.62962963
= 55.17037037 m/s
speed of second stone
= v + at
= 25+9.8*3.62962963
= 60.57037037 m/s
