Solution is here,
for initial case,
temperature(T1)=70°C=70+ 273=343K
vloume( V1) =45 L
for final case,
temperature( T2)=?
volume(V2)= 91.3 L
at constant pressure,
V1/V2 = T1/T2
or, 45/91.3 = 343/ T2
or, T2= (343×91.3)/45
or, T2=695.9 K = (695.9-273)°C=422.9°C
Answer:
Explanation:
We are given the mass, specific heat, and temperature, so we must use this formula for heat energy.
The mass is 5 grams, the specific heat capacity is 0.14 Joules per gram degree Celsius. Let's find the change in temperature.
- ΔT= final temperature - initial temperature
- ΔT= 95°C - 15°C = 80°C
We know the variables and can substitute them into the formula.
Multiply the first numbers. The grams will cancel.
Multiply again. This time the degrees Celsius cancel.
56 Joules of heat are needed.
Answer:
The parts of an atom are<em><u> protons, electrons, and neutrons.</u></em>
A proton is positively charged and is located in the center or nucleus of the atom.
Electrons are negatively charged and are located in rings or orbits spinning around the nucleus.
The number of protons and electrons is always equal.
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
The answer is 6.02e24 molecules or 6.02x10^24 molecules.
