Your answer would be a change in odor! Hope this helps! ;D
Answer:
126.0g of water were initially present
Explanation:
The electrolysis of water occurs as follows:
2H₂O(l) ⇄ 2H₂(g) + O₂(g)
<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>
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To find the mass of water we need to determine moles of oxygen and hydrogen, thus:
<em>Moles Hydrogen:</em>
14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂
<em>Moles Oxygen:</em>
112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂
Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:
7 moles H₂O * (18g / mol) =
<h3>126.0g of water were initially present</h3>
Answer:
Keq =1.50108
Explanation:
The given reactionis
C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)
ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )
= -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)
= - 242.09kJ/mol
ΔG= -RTlnKeq
ln Keq = -ΔG/RT
=-(- 242.09kJ/mol ) / 2 k cal /mol*298 K
=0.406
Keq =e0.406
Keq =1.50108
Answer: D.) 25.9%
Explanation:
Dinitrogen pentoxide chemical formular : N2O5
Calculating the molar mass of N2O5
Atomic mass of nitrogen(N) = 14
Atomic mass of oxygen(O) = 16
Therefore molar mass :
N2O5 = (2 × 14) + (5 × 16) = 28 + 80 = 108g/mol
Percentage amount of elements in N205:
NITROGEN (N) :
(Mass of nitrogen / molar mass of N2O5) × 100%
MASS OF NITROGEN = (N2) = 2 × 14 = 28
PERCENT OF NITROGEN : (28/108) × 100%
0.259259 × 100%
= 25.925%
= 25.9%
