<h3><u>Answer;</u></h3>
Higher velocity of particles
<h3><u>Explanation;</u></h3>
The diffusion rate is determined by a variety of factors which includes;
- Temperature such that the higher the temperature, the more kinetic energy the particles will have, so they will move and mix more quickly and the diffusion rate will be high.
- Concentration gradient such that the greater the difference in concentration, the quicker the rate of diffusion.
- Higher velocity of particles increases the diffusion rate as this means more kinetic energy by the particles and hence the particles will mix and move faster, thus higher diffusion rate.
Answer:
59.92 × 10²³ atoms are in 9.95 moles of iron
1.8 ×10²² molecules are in 0.03 moles of Carbon dioxide
1.19 moles are found in 7.20 x 10^23 atoms of platinum
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
<span>Higher energy = shorter wavelength
Frequency is one cycle over an amount of time (seconds)
So higher frequency = higher energy = shorter wavelength</span>
