Answer:
The answer is 12,560
Explanation:
The orbital period is the time a given cosmic question takes to finish one circle around another protest and applies in space science as a rule to planets or space rocks circling the Sun, moons circling planets, exoplanets circling different stars, or double stars. Mercury, for instance, has an orbital time of 88 days while it takes Jupiter around 11.86 years. The time of the Earth's circle is generally thought to be 365 days as timetables appear.
Answer:
Q = 1267720 J
Explanation:
∴ QH2O = mCpΔT
∴ m H2O = 500 g
∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C
∴ ΔT = 120 - 50 = 70°C
⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ
∴ ΔHv H2O = 40.7 KJ/mol
moles H2O:
∴ mm H2O = 18.015 g/mol
⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O
⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ
⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J
Answer:
Explanation:
We have the equation for ideal gas expressed as:
PV=nRT
Being:
P = Pressure
V = Volume
n = molar number
R = Universal gas constant
T = Temperature
From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:
So
Now, as for the final condition:
clearing
Answer:
32.23 to 4 significant figures.
Explanation:
The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).
So by proportion it is 6.022 * 10^23 * 3.88 / 7.25 * 10^22
= 32.23 to 4 significant figures.
Answer:
Explanation:
Hello,
In this case, since the 3% by volume is represented as:
By using the ideal gas equation we compute the density of CO:
Then we apply the conversion factors as follows:
Regards.
