The heat lost by the metal should be equal to the heat
gained by the water. We know that the heat capacity of water is simply 4.186 J
/ g °C. Therefore:
100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp *
(95.2°C - 31°C)
<span>Cp = 1.36 J / g °C</span>
Answer:
56972.17K
Explanation:
P = 4.06kPa = 4.06×10³Pa
V = 14L
n = 0.12 moles
R = 8.314J/Mol.K
T = ?
We need ideal gas equation to solve this question
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
T = PV / nR
T = (4.06×10³ × 14) / (0.12 × 8.314)
T = 56840 / 0.99768
T = 56972.17K
Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa
Answer:The environmental problems directly related to energy production and Electricity is a secondary energy source that is generated producedfrom primary energy sources:
hope that helps im new so have a good day (:
Explanation:
Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
Answer:
The answer to your question is 3% H2SO4 solution
Explanation:
Data
Concentration 2 = C₂ = ?
Concentration 1 = C₁ = 15 %
Volume 1 = V₁ = 50 ml
Volume 2 = V₂ = 250 ml
Formula
C₁V₁ = C₂V₂
Solve for C₂
C₂ = C₁V₁ / V₂
Substitution
C₂ = (15)(50) / 250
Simplification and result
C₂ = 3 %
