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3 years ago

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

Chemistry

1 answer:

Rashid [163]3 years ago

5 0

Answer:

4.06 mol H₂O

Explanation:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

First we convert the given masses of reactants into moles, using their respective molar masses:

250 g O₂ ÷ 32 g/mol = 7.81 mol O₂

50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄

Now we calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles, using the stoichiometric coefficients of the reaction:

0.58 mol C₆H₁₄ * $\frac{19molO_2}{2molC_6H_{14}}$ = 5.51 mol O₂

As there are more O₂ moles than required (7.81 vs 5.51), O₂ is the reactant in excess. That means that C₆H₁₄ is the limiting reactant.

Now we can calculate how much water can be formed, using the number of moles of the limiting reactant:

0.58 mol C₆H₁₄ * $\frac{14molH_2O}{2molC_6H_{14}}$ = 4.06 mol H₂O

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Explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

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$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

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Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

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3 years ago

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

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Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

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Therefore, the pH of a saturated solution is, 12.33

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