Answer:
The answer to your question is: letter c (96%)
Explanation:
Indium -113 (112-9040580 amu) ₁₁₃In
Indium-115 (114.9038780 amu) ₁₁₅In
Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In
Formula
Atomic mass = m₁(%₁) +m₂(%₂) / 100
%₁ = x I established this is an equation
%₂ = 100 - x
Substituting values
114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify
114.82 = 112.8040x + 11490.39 - 114.9039x /100
11482 = 112.8040x -114.9039x +11490.39
11482 - 11490.39 = 112.8040x -114.9039x
-8.39 = -2.099x
x = 3.99
Then % of Indium-115 = 100 - 3.99 = 96
Answer:1. Energy 2. Medium 3. Original position 4. Waves 5. Matter
Explanation:
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
<h2>Answer:</h2>
The correct answer is the option A which is: Cl2 + 2e- → 2Cl-
<h3>Explanation:</h3>
<em><u>Reduction is the gain of electrons and resulting in neutral or negative ions.</u></em>
<em><u>It is also the gain of hydrogen and release of oxygen ions.</u></em>
- According to first most definition, option A describes the reduction.
- Option B is incomplete reaction.
- While C and D are oxidation reactions.
Bing bling or expansive jewellery
