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
Answer:
Any element placed in a flame will change its color. Atoms are made of positively charged nuclei, about which negatively charged electrons move according to the laws of quantum mechanics. Quantum mechanics constrains them to appear in various distinct patterns, called orbitals. (Orbitals are a lot like planetary orbits, but blurrier, so that you're never quite sure just where the electrons are.)
Left on their own, the electrons of an atom tend to relax into orbitals that leave the atom with the lowest possible energy--its ground state. Putting atoms into a flame, though, adds energy to the looser electrons farthest from the nucleus and pushes them into other orbitals. Eventually, these excited electrons drop back to where they ought to be, and in so doing, they release the energy they stored up as particles of light, called photons.
Explanation:
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