2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
<em>Step 1</em>. Write the <em>condensed structural formula</em> for 2,3-dimethylbutane.
(CH_3)_2CHCH(CH_3)_2
<em>Step 2</em>. Write the <em>molecular formula</em>.
C_6H_14
<em>Step 3</em>. Write the <em>unbalanced chemical equation</em>.
C_6H_14 + O_2 → CO_2 + H_2O
<em>Step 4</em>. Pick the <em>most complicated-looking formula</em> (C_6H_14) and balance its atoms (C and H).
<em>1</em>C_6H_14 + O_2 → <em>6</em>CO_2 + <em>7</em>H_2O
<em>Step 5</em>. Balance the <em>remaining atoms</em> (O).
1C_6H_14 + (<em>19/2</em>)O_2 → 6CO_2 + 7H_2O
Oops! <em>Fractional coefficients</em>!
<em>Step 6</em>. <em>Multiply all coefficients by a number</em> (2) to give integer coeficients..
2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
Answer:
number of carbon-carbon single (C - C) bonds: 1
number of carbon-hydrogen single (C H) bonds: 5
number of nitrogen-hydrogen sing le (N H) bonds:2
number of lone pairs: 1
Explanation:
Ethanamine is a colourless gas having a strong 'ammonia- like' odour. It contains the -NH2 group which makes it an amine. It contains one carbon-carbon bond, five carbon-hydrogen bonds and two nitrogen-hydrogen bonds.
Nitrogen, being sp3 hybridized in the compound has a lone pair of electrons localized on one of the sp3 hybridized orbitals of nitrogen while one sp3 hybridized orbital of nitrogen is used to form a carbon-nitrogen bond. The other two sp3 hybridized orbitals on nitrogen are used to form the two nitrogen-hydrogen bonds.
Answer:
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<span>By definition, the first ionization energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+. </span><span />
Answer:
537.68 torr.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and V are constant, and have different values of P and T:
<em>(P₁T₂) = (P₂T₁).</em>
P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,
P₂ = ??? torr, T₂ = 74°C + 273 = 347 K.
∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.
