Answer: A) The reaction is spontaneous above 276 K.
Explanation:
According to Gibb's equation:
= Gibbs free energy
= enthalpy change = +29.3 kJ/mol =29300 J/mol
= entropy change = +106 J/molK
T = temperature in Kelvin
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
for reaction to be spontaneous
Thus the Reaction is spontaneous when temperature is above 276 K.
Answer:
1. 17 protons / 10 neutrons
2. Scandium
3. Negative
Explanation:
1. The bottom number is the same as the atomic number on top of the abbreviation. To find the # of neutrons, subtract the mass number with the number of protons
2. The bottom number (21) is the number of protons AND the atomic number (the a.n & # of pro. are always going to be the same)
3. Always think about this: if you add electrons, you take away charge - if you take away electrons, you add charge
*Idek if these were the correct correct answers but I tried my best*
Answer:
CO.
Explanation:
Assuming the given percentages are by mass, we can solve this problem via imagining we have <em>100 g of the compound</em>, if that were the case we would have:
Now we <u>convert those masses into moles</u>, using the<em> elements' respective molar masses</em>:
- 42.9 g of C ÷ 12 g/mol = 3.57 mol C
- 57.1 g of O ÷ 16 g/mol = 3.58 mol O
As the number of C moles and O moles is roughly the same, the empirical formula for the compound is <em>CO</em>.
Answer:
0.200M H₃PO₄
0.600N H₃PO₄
pH = 1.46
Explanation:
The acid-base reaction of phosphoric acid (H₃PO₄) with LiOH is:
3 LiOH + H₃PO₄ → Li₃PO₄ + 3H₂O
<em>Where 3 moles of LiOH reacts per mole of H₃PO₄</em>
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Moles of LiOH are:
0.030L× (0.5mol / L) = 0.0150 moles of LiOH
Moles of acid neutralized are:
0.0150 moles of LiOH × (1 mole H₃PO₄ / 3 moles LiOH) = 0.005 moles H₃PO₄
As volume of acid was 25mL, molarity is:
0.005mol H₃PO₄ / 0.025L =<em> 0.200M H₃PO₄</em>
Normality is:
0.200M × (3N H⁺ / 1M H₃PO₄) = <em>0.600N H₃PO₄</em>
H₃PO₄ dissolves in water thus:
H₃PO₄ ⇄ H₂PO₄⁻ + H⁺
Ka = 7.1x10⁻³ = [H₂PO₄⁻] [H⁺] / [H₃PO₄]
Where molar concentrations in equilibrium will be:
[H₂PO₄⁻] = X
[H⁺] = X
[H₃PO₄] = 0.200M - X
Replacing in Ka formula:
7.1x10⁻³ = [X] [X] / [0.200 - X]
1.42x10⁻³ - 7.1x10⁻³X = X²
0 = X² + 7.1x10⁻³X - 1.42x10⁻³
Solving for X:
X = -0.04M →False answer, there is no negative concentrations.
X = 0.0343M
As, [H⁺] = 0.0343M
pH = - log [H⁺],
<em>pH = 1.46</em>
Dots are used to represent electrons.
