Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>
Answer:
22.8 L
Step-by-step explanation:
We can use <em>Gay-Lussac's Law of Combining Volumes</em> to solve this problem:
Gases <em>at the same temperature and pressure</em> react in simple whole-number ratios.
1. Write the chemical equation.
Ratio: 2 L 1 L
Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)
V/L: 11.4
2. Calculate the volume of HCl.
According to the law, 2 L of HCl form 1 L of H₂.
Then, the conversion factor is (2 L HCl/1 L H₂).
Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)
= 22.8 L HCl
Answer:
You will get 5.0 g of hydrogen.
Explanation:
As with any stoichiometry problem, we start with the balanced equation.
Sn
l
+
2HF
→
SnF
2
+
H
2
Moles of H
2
=
2.5
mol Sn
×
1 mol H
2
1
mol Sn
=
2.5 mol H
2
Mass of H
2
=
2.5
mol H
2
×
2.016 g H
2
1
mol H
2
=
5.0 g H
2
Hybridization in ozone, O3......
<span>...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O </span>
<span>../ </span>
<span>O .................. 3 lone pairs on terminal O </span>
<span>I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. </span>
<span>The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. </span>
<span>If the electron pair geometry is linear, the hybridization is sp. </span>
<span>If the electron pair geometry is trigonal planar, the hybridization is sp2. </span>
<span>If the electron pair geometry is tetrahedral, the hybridization is sp3. </span>
<span>The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.</span>
