Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Bromine vs Chlorine | Br vs Cl
Halogens are group VII elements in the periodic table, and all are electronegative elements and have the capability to produce -1 anions.
Bromine
Bromine is denoted by the symbol Br. This is in the 4th period of the periodic table between chlorine and iodine halogens. Its electronic configuration is [Ar] 4s2 3d10 4p5. The atomic number of bromine is 35. Its atomic mass is 79.904. Bromine staChlorine is an element in the periodic table which is denoted by Cl. It is a halogen (17th group) in the 3rd period of the periodic table. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Its electron configuration is written as 1s2 2s2 2p6 3s2 3p5. Since the p sub level should have 6 electrons to obtain the Argon, noble gas electron configuration, chlorine has the ability to attract an electron. ys as a red-brown color liquid at room temperature.
Answer:
4 valence electrons
Despite the fact that the word silicon has a ubiquitous affiliation with all things electronic, Si itself is not a good electrical conductor. It has 4 valence electrons, meaning that filling its outer shell it can form a very strong lattice with 4 neighboring Si atoms-with no un-bonded electrons remaining
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:
Different compounds react with oxygen differently – some contain lots of heat energy while others produce a smaller amount. The reaction with the oxygen may happen very quickly or more slowly. Amount: The amount of fuel available to burn is known as the fuel load.
Explanation:
