Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
They are considered malleable. They can be made into sheets
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Answer:
Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.
<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>
Explanation:
Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.
Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.
Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.
The final temperature of the water is the equilibrium temperature, or the also the final temperature of the iron after a long period of time. Applying the conservation of energy:
m,iron*C,iron*ΔT = - m,water*C,water*ΔT
The density of water is 1000 g/mL.
(25 g)(0.449 J/g·°C)(T - 398 K) = - (25 mL)(1000 g/mL)(4.18 J/g·°C)(T - 298)
Solving for T,
<em>T = 298.01 K</em>
