Answer:
1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. V = 596L
Explanation:
Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:
C₄H₁₀ + O₂ → CO₂ + H₂O
1. The balanced chemical equation is:
C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. 0,360kg of butane are:
360g×
=<em>6,19moles of butane</em>
These moles of butane are:
6,19moles of butane×
= <em>24,8 moles CO₂</em>
Using V=nRT/P
Where:
n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).
Volume (V) is:
<em>V = 596L</em>
I hope it helps!
You'll want to add three amounts of heat.
(1) Specific heat of lowering the temperature from -135°C to the melting point -114°C
(2) Latent heat of fusion/melting
(3) Specific heat of elevating the temperature from -114°C to -50°C
(1) E = mCΔT = (25 g)(0.97 J/g·°C)(1 kJ/1000 J)(-114 - -135) = 0.509 kJ
(2) E = mΔH = (25 g)(5.02 kJ/mol)(1 mol/46.07 g ethanol) = 2.724 kJ
(3) E = mCΔT = (25 g)(2.3 J/g·°C)(1 kJ/1000 J)(-50 - -114) = 3.68 kJ
<em>Summing up all energies, the answer is 6.913 kJ.</em>
Yes. Mercury has 80 protons. Tin has 50 protons. Same for electrons, it just doesn't have an exact number.
Part 1)
Cu- <span>[Ar] 3d¹⁰4s¹ </span><span>atomic number: 29
</span>
<span>O- [He] 2s2 2p<span>4 atomic number:8
</span></span>La- <span>[Xe] 5d¹ 6s² </span><span>atomic number:57
Y- </span><span>[Kr] 4d¹5s² </span><span>atomic number:39
Ba- </span><span>[Xe] 6s² </span><span>atomic number:56
Tl- </span><span>[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹ </span><span>atomic number:81
Bi- </span> <span>[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p³ </span>atomic number:83
Part 2)
You are able to this by consulting the periodic table and following this steps:
-Find your atom's atomic number;
<span>-Determine the charge of the atom (these were all uncharged)
</span><span>-Memorize the order of orbitals (s, d, p, d.. and how many electrons they can fit)
</span>-<span>Fill in the orbitals according to the number of electrons in the atom
- </span><span>for long electron configurations, abbreviate with the noble gases</span>
