<span>The correct answer should be two oxygen atoms. That's because it's properties are similar to carbon insofar that it can form four bonds, so if it forms bonds with 2 oxygen atoms then it will have all four bonds created since Oxygen forms double bonds. This would make SiO2 which is also known worldwide as silica.</span>
As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
Answer:
17.57kg of 
and its percentage yield is 81.0%
Explanation:
Through the reaction you can get the theoretical amount of that must be produced.
If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage
=81.0%
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
Answer: 19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2
