The equation for the reaction of hydrogen and oxygen is:
![2H_2 + O_2 \rightarrow 2H_2O](https://tex.z-dn.net/?f=2H_2%20%2B%20O_2%20%5Crightarrow%202H_2O)
![M_r(H_2) = 2\\\\\therefore n(H_2) = \frac{1}{2}= 0.5mol = n(H_2O)\\\\m(H_2O) = 0.5 \times M(H_2O) = 0.5 \times 18 = 9g](https://tex.z-dn.net/?f=M_r%28H_2%29%20%3D%202%5C%5C%5C%5C%5Ctherefore%20n%28H_2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%3D%200.5mol%20%3D%20n%28H_2O%29%5C%5C%5C%5Cm%28H_2O%29%20%3D%200.5%20%5Ctimes%20M%28H_2O%29%20%3D%200.5%20%5Ctimes%2018%20%3D%209g)
Thus, the theoretical yield of water is 9 grams.
Answer:
Ni^2+ is most likely
Ti^3+ is very unlikely
Explanation:
The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.
Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.
<h3>
Answer:</h3>
0.0113 mol Ba(ClO₃)₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structures</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.45 g Ba(ClO₃)₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ba - 137.33 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Ba(ClO₃)₂ - 137.33 + 2(35.45) + 6(16.00) = 304.33 g/mol
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 3.45 \ g \ Ba(ClO_3)_2(\frac{1 \ mol \ Ba(ClO_3)_2}{304.33 \ g \ Ba(ClO_3)_2})](https://tex.z-dn.net/?f=%5Cdisplaystyle%203.45%20%5C%20g%20%5C%20Ba%28ClO_3%29_2%28%5Cfrac%7B1%20%5C%20mol%20%5C%20Ba%28ClO_3%29_2%7D%7B304.33%20%5C%20g%20%5C%20Ba%28ClO_3%29_2%7D%29)
- Multiply/Divide:
![\displaystyle 0.011336 \ mol \ Ba(ClO_3)_2](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.011336%20%5C%20mol%20%5C%20Ba%28ClO_3%29_2)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.011336 mol Ba(ClO₃)₂ ≈ 0.0113 mol Ba(ClO₃)₂