So if the sum of the enthalpies of the reactants is greater than the products, the reaction will be exothermic. If the products side has a larger enthalpy, the reaction is endothermic. You may wonder why endothermic reactions, which soak up energy or enthalpy from the environment, even happen
<span>Answer: 17.8 cm
</span>
<span>Explanation:
</span>
<span>1) Since temperature is constant, you use Boyle's law:
</span>
<span>PV = constant => P₁V₁ = P₂V₂
</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:
</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
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<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:
</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³
</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>
Answer:
30 g. 8'C
Explanation:
the 10 g is plus the fact that I have to help you get the chance to get the answer
F. <em>None of the above
</em>
<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.
O_2 + 2H_2O + 2e^(-) → 4OH^(-)
We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.
The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).
A decrease in oxidation number is <em>reduction</em>, so O_2 is the substance reduced.
The oxidation number of O is -2 in both H_2O and OH^(-), so water is <em>neither oxidized nor reduced</em>.
Answer:
CaO is a binary compound.
