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:
6.4 L
Explanation:
When all other variables are held constant, you can use Boyle's Law to find the missing volume:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.
P₁ = 3.2 atm P₂ = 1.0 atm
V₁ = 2.0 L V₂ = ? L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.2 atm)(2.0 L) = (1.0 atm)V₂ <----- Insert values
6.4 = (1.0 atm)V₂ <----- Simplify left side
6.4 = V₂ <----- Divide both sides by 1.0
Answer:
The temperature associated with this radiation is 0.014K.
Explanation:
If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.
where,
λmax is the wavelength at the peak of emission
b is Wien's displacement constant (2.89×10⁻³ m⋅K)
T is the absolute temperature
For a wavelength of 21 cm,
